Calculus and Analytic Geometry II Chapter - Sequences and Series. Sequences Definition. A sequence is a list of numbers written in a definite order, We call a n the general term of the sequence. {a, a 2, a 3,...} = {a n }. Example. Assuming that the pattern of the first few terms continues, find a formula for the general term a n of each sequence. () (2) (3) {, 2, 3, 4 } { },... = n { 4, 29, 36, 425 } {,... = ( ) n n (n + ) 2 { 4, 2 9, 3 6, 4 } { 25,... = ( ) n+ n (n + ) 2 Remark. Some sequences (actually, many sequences) do not have a simple formula. Consider the Fibonacci sequence {,, 2, 3, 5, 8, 3, 2,...}. } } This is an example of a recursive sequence. There is a closed formula expression for the nth Fibonnaci number F (n) but it is harder to write down than the previous problem. Let φ = 2 ( + 5) (the golden ratio) and ψ = 2 ( 5). Then F (n) = φn ψ n 5. Definition 2 (Informal). A sequence {a n } has the it L if we can make the terms a n as close to L as we like by taking n sufficiently large. We write a n = L or a n L as n. If the it exists, we say the sequences converges. Otherwise, we say the sequence diverges. Definition 3 (Formal). A sequence {a n } has the it L if for every ε > 0 there is a corresponding number N > 0 such that a n L < ε for all n > N.
Definition 4. The expression a n = means that for every positive number M there is an integer N such that a n > M for all n > N. Example 2. Show that n = 0. Let ε > 0. We must find a number N such that n 0 < ε. This is equivalent to < ε. Solving for n gives n >. Set N =. Then if n > N then n ε ε < ε. n Example 3. Show that ln(n) =. Let M > 0. To find N, set ln(n) > M. This is equivalent to n > e M. Set N = e M. Now if n > N we have ln(n) > M. Remark. Note that we can plot these points on a graph using the ordered pairs (n, a n ). If our sequence is determined by a function, then the long term behavior of the sequence is identical to that of the function. The following theorem formalizes that concept. Definition 5. Let {a n } be a sequence. If f is a function such that f(n) = a n for each integer n, then we say the function f models the sequence {a n }. Theorem 4. If f is a function which models the sequence {a n } and then f(x) = L x a n = L. Remark. Note that the above theorem does not apply when f diverges. Consider the sequence {sin(πn)}. This sequence converges to 0, in fact, every term in the sequence is 0. However, while f(x) = sin(πx) does model {a n }, the it of f(x) as x does not exist. 2
Example 5. Use the previous theorem to evaluate the it of the sequence Let f(x) = x. Now, x 2 + f(x) = x x a n = n n2 +. x x x2 + = 2 x x 2 + = x x 2 x 2 + = =. Theorem 6. Let a n = p(n) where p(n) and q(n) are polynomials in n with leading coefficients q(n) b and c, respectively. Then 0 if deg q(n) > deg p(n) n = if deg p(n) > deg q(n) Example 7. Find the it of the sequence, { } n!. (n + 2)! Notice that we could rewrite the general term b c if deg p(n) = deg q(n). n! (n + 2)! = n! (n + 2)(n + )n! = (n + 2)(n + ) 0. Remark. We ll now discuss several theorems that help us to evaluate its of sequences. Many of these should remind you of corresponding theorems for its of functions. The last one relates the two types of its. Theorem 8 (Limit Laws for sequences). If {a n } and {b n } are convergent sequences and c is a constant, then () (a n ± b n ) = a n ± b n (2) ca n = c a n and c = c (3) (a n b n ) = a n b n a n (4) = a n b n b n if b n 0 Theorem 9 (Squeeze Theorem for Sequences). Let {a n }, {b n }, {c n } be sequences such that for some number M, b n a n c n for n > M and b n = c n = L, 3
then a n = L. Example 0. Evaluate the it of the sequence with general term a n = / n 4 + n 8. We can bound a n by 2n 4 a n 2n 2. Each of these sequences converges to 0 and then by the Squeeze Theorem, so does {a n }. Theorem. If a n = 0, then a n = 0. Proof. We have a n a n a n. Since a n = 0, then so too do we have a n = 0. Hence, by the Squeeze Theorem, a n = 0. Remark. The previous theorem only works if the it is 0. {( ) n } = {,,,,,,...}. sequence diverges. Theorem 2. If f(x) is continuous and then Consider the sequence Then the it of the absolute values is but the a n = L ( ) f(a n) = f a n = f(l). Remark. The sequence {r n } is convergent if < r and otherwise it is divergent. 0 < r < rn = r = r >. 2. Series Remark. Consider the process of adding all of the integers. We could do this by hand, + 2 + 3 +, but that would take a while, and we re pretty sure its infinity. Thankfully, there is a formula that tells us the sum of the first n integers S n = n(n + ), 2 4
and we can prove this with induction. Now we just increase n to get bigger and bigger sums. The total sum is just the it of these partial sums, so an = S n =. Now let s consider a similar problem, the sequence given by {/2 n }. The more successive terms we add, the closer we get to. Hence, we might guess that 2 n =. We can see this geometrically by drawing a square, cutting it in half and then cutting the remaining half in half... This forms a sequence, { 2, 3 4, 7 8, 5 } 6,.... Hence, a general term in the sequence has the form This sequence clearly converges to. 2 n 2 n = 2 n. Definition 6. A series (or infinite series) is the infinite sum of the terms in a sequence {a n }. We denote it a n or a n. Let S N denote its Nth partial sum, that is, S N = N a n = a + a 2 + + a N. If the sequence {S N } is convergent and N s N convergent and write an = S. = S exists, we say the series a n is The number S is called the sum of the series. If the sequence {S n } is divergent, then the series is called divergent. Remark. We will often use the notation, n a n = a i. We can add and subtract series termwise (see page 55). Moreover, a scalar multiple of an infinite series is the same as the infinite series of the terms multiplied by that scalar. 5 i=
Theorem 3. The geometric series is convergent if r < and its sum is cr n cr n = If r, the geometric series diverges. c r. Proof. If r = then the series clearly diverges. If r, then observe that S N = c + cr + cr 2 + + cr N rs N = cr + cr 2 + + cr N S N rs N = c cr N S N ( r) = c( r N ). Thus, ( c ) r N S = S N = r = c r c r rn. Thus, the series converges if r < and otherwise it diverges. Example 4. Determine whether each series converges or diverges. If it is convergent, find its sum. () 3 2 + 9 4 27 8 + This is a geometric series with r = 3/2. Since r >, then this series diverges. (2) 4 n+ 5 n We rewrite the series as 4 n+ 5 n = 4 + 4 n+ 5 n = 4 + = 4 + 6/5 (4/5) = 4 + 6 = 20. 6 4 2 4 n 5 5 n = 4 + 6 5 ( ) n 4 5
Example 5. Express the number 6.254 = 6.254545454 as a ratio of integers. We can write this number as the series as 6.254 = 6.2 + 54 000 + 54 54 + = 6.2 + 00000 0 + 54 3 0 + 54 5 0 + 7 = 6.2 + 54 ( + 0 3 0 + ) ( 2 0 + = 6.2 + 54 + 4 0 3 00 + = 6.2 + ( ) n 54. 0 3 00 ( ) ) 2 + 00 This is a geometric series with c = 54/0 3 and r = 5400 < 0. Hence, the series converges to 54/03 = 54 = 54 = 6 62. Thus, the value of the number is + 6 = 344. /00 000 0 990 0 0 0 55 Example 6. The series is known as the harmonic series. The harmonic series n diverges. We will show why below. Instead of looking for each partial sum, we will bound some of them. We use the following fact: If {a n } is a sequence containing a subsequence {b n } which diverges, then {a n } diverges. We have s 2 = + 2 ( ) s 4 = + 2 + 3 + 4 > + 2 4 + = + 2 4 2 s 8 = + ( 2 + 3 + ) ( + 4 5 + 6 + 7 + ) 8 > + ( 2 + 4 + ) ( + 4 8 + 8 + 8 + ) = + 3 8 2. This continues and we find s 2 n > + n 2 and so s 2 n as n. Hence {s n} diverges. Example 7. Determine the convergence of the series n(n + ). If it converges, find its sum. First note that. And so the given series should converge if n+ n 2 converges. n 2 (We have not yet formalized a comparison theorem for series, but it should be rather intuitive.) This series should remind us of the indefinite integral Hence, our intuition should tells us that the given series converges. 7 x 2 dx, which converges.
We use the identity n(n + ) = n n +, which comes from partial fraction decomposition. Hence, ( S N = ) ( + 2 2 ) ( + 3 3 ) ( + + 4 N ) N Since then the sum of the series is. S N =, N = N +. Remark. The series in Example 7 is known as a telescoping series. The strategy for finding the sum of any telescoping series is the same as in that example. Theorem 8. If the series a n is convergent, then a n = 0. Theorem 9 (Divergence Test). If a n 0, then the series a n diverges. Example 20. Show that the following series diverges n 0n + 2. Test. Let a n = n 0n+2, then a n 0 as n. Hence, the series diverges by the Divergence Remark. Note that the converse of the Divergence Test does not hold. There are sequences such that the it of the terms tends to zero but the series does not converge, e.g. the harmonic series. 3. The Integral Test Remark. We now turn to the question of when certain series converge or diverge. In this section and the next, we will focus exclusively on series of the form a n with a n 0 for all n. Remark. Consider the series n. We treat each summand as the area of a rectangle with base and height n. All of these rectangles can be position so that they lie above the curve x. Thus,. n x 8
Hence, the sum diverges. Now consider the series. Starting with n = 2, we proceed as above by treating n 2 each summand as a rectangle with base and height. then we see that these rectangles n 2 lie below the curve x 2. Hence, it is easy to see that n=2 n 2 x. 2 Since this series converges, it has a finite sum. Thus, the sum of the original series is plus this sum, and is stil infinite. Theorem 2 (The Integral Test). Suppose f(x) is continuous, positive, and decreasing function on [, ) which models the sequence {a n }. () If (2) If f(x) dx converges, then so does f(x) dx diverges, then so does a n, a n. Example 22. Determine whether the following series converges or diverges. We will consider the integral, ln n n 2. and apply IBP. Let u = ln x and dv = x 2 dx so du = x dx and v = x. Then ( ln x b [ ln x dx = dx = ln x ] b ) b + x 2 b x 2 b x x dx 2 ( = ln b [ b b + 0 + ] ) b = 0 + [ b ] x + =. b ln x x 2 Hence, by the integral test, the given series converges. dx Caution! The above argument does not tell us that the sum of the series is. It only tells us that the series converges. The actual value of the series is a bit less than (approx.9375) but this is difficult to determine. 9
Example 23. Determine whether each series converges or diverges. () Let f(x) = x /4. Then f models the sequence { 4 n. 4 n }. Since dx x/4 diverges (p-integral type I, p = /4), then so does the given series. (2) Let f(x) = 3 5x 4. Then f models the sequence { 3 3 5n 4. 5n 4 }. Since 3 5x dx = 3 4 5 x dx 4 converges (p-integral type I, p = 4), the so does the given series. Theorem 24 (p-test for series). The infinite series converges if p > and diverges otherwise. n p 4. Comparison Tests Remark. In this section we will consider two different comparison tests. The first should seem very familiar. Theorem 25 (The Comparison Test for Series). Assume there exists M > 0 such that 0 a n b n for n M. If b n converges, then a n also converges. If a n diverges, then b n also diverges. Remark. Why do we care about n M in the above theorem and don t just say for all n? The point is that it only matters what happens eventually in the sequence, not what happens in the early terms. 0
Example 26. Determine whether each series converges or diverges. ln n () n. For n > e, ln(n) >. Since the harmonic series diverges, then so too does our given series n n by the Comparison Test. (2) n 2 + n +. We have n 2 + n + > n 2 for n > 0. Hence, <. Since the series n 2 +n+ n 2 n 2 the p-test, p >, then by the Comparison Test so too does the given series. (3) n2 +. converges by Since n 2 + > n 2 = n, then < n. Thus, our series is smaller than a divergent 2 + n series and so the Comparison Test does not apply. Remark. The Comparison Test gives us a way of comparing series in which the terms in the series are bigger or smaller than those in a convergent or divergent series. The next test is similar, but compares growth rates of series. Theorem 27 (The Limit Comparison Test). Suppose that a n and b n are series with positive terms. If a n = c b n where c is a finite number with c > 0, then either both series converge or both diverge. Example 28. Use the Limit Comparison test to determine whether the following series converges or diverges n2 +. We compare to the harmonic series, which has terms b n = /n. / n 2 + n n = /n n2 + = 2 n 2 + = n 2 n 2 + = > 0. Thus, by the Limit Comparison Test, the given series diverges. Example 29. Determine whether the following series converges or diverges. cos 2 (3n) + (.2). n
First note that 0 cos 2 (3n). Hence, cos 2 (3n) + (.2) n + (.2) n (.2). n The sequence is geometric (r = 5 (.2) n 6 Comparison Test, the given series converges. so r < ) and so it converges. Thus, by the Example 30. Determine whether the following series converges or diverges. 3n 3 + 2n 5n 5 2n 3 + 3. It would be difficult to apply the Comparison Test (though not impossible perhaps). We will instead apply the Limit Comparison Test. Set a n = 3n3 +2n. To determine the 5n 5 2n 3 +3 comparing series, look for the highest power of n in the numerator and denominator. We will compare to the series b n where b n = n3 =. By the p-test, this series converges n 5 n 2 a (p > ). Thus, if n bn = c where c > 0 and finite, then the given series converges. a n (3n 3 + 2n )/(5n 5 2n 3 + 3) = b n (/n 2 ) Hence, the given series converges by the Limit Comparison Test. = 3n 5 + 2n 3 n 2 5n 5 2n 3 + 3 = 3 5 > 0. Remark. The next example illustrates that there may be times when it is advantageous to use both theorems in conjunction. Example 3. Show that the series converges. n! Since n! grows fast, we will conjecture that the series converges. For n > we have n! > n(n ) so n! < n(n ). By the Comparison Test, our given series will converge if the series n(n ) converges. We will use the Limit Comparison Test. Let a n = n(n ) and b n = n 2. Then Thus, by the LCT, n(n ) converges. a n n 2 = b n n(n ) =. Remark. We will see that there is a much quicker way to show convergence of this series in Section 6. 2
5. Alternating series Definition 7. An alternating series is a series whose terms are alternately positive and negative. Example 32. Consider the alternating harmonic series ( ) n n, or its variation ( ) n n, which is just the additive inverse of the one before. We saw previously that the alternating harmonic series converges. It turns out that the sum is ln(2), though the reason for this will come later. (If you want a hint, start thinking about the derivatives of ln(x). Remark. What we will consider in this section is a test for convergence of alternating series. Theorem 33 (The Alternating Series Test). If the alternating series, ( ) n b n = b b 2 + b 3 b 4 + b 5 b 6 + b n > 0 satisfies then the series is convergent. (i) b n+ b n for all n (decreasing) and (ii) b n = 0. Remark. The Alternating Series Test works similarly if ( ) n is replaced by ( ) n. It is also not affected by the starting value. Example 34. Use the alternating series test to give another explanation of why the alternating harmonic series converges. We need to check two things: that the series is decreasing and the it of the summands in zero. The second is obvious, n = 0. For the first, we need to show for all n. This is clear because the denominator on n+ n the left is always larger. Equivalently, we could cross multiply to be n n +, or 0. Hence, by the Alternating Series Test, the series converges. 3
Remark. Why does this test work? We give a weak argument based on observation. A more complete argument is in your textbook. Start with b, drawn out on a line segment. We then subtract from that b 2, which is less than b by hypothesis. Next, we add on b 3, which is less than b 2 and subtract b 4. This difference is completely contained in b 2. Continuing in this process, we see that the it of partial sums is bounded by b. Hence, the sequence of partial sums is increasing (monotonic) and bounded, therefore convergent. What issues are this argument ignoring? Remark. Given a series as in the theorem above, we have 0 < S < b and S 2N < S < S 2N+. Example 35. Determine the convergence of the series ( ) n n2 +. This is similar to the above. Note that the fact that the powers of are off from the test does not affect the convergence. Example 36. Determine the convergence of the series cos(πn). n! This is an alternating series in disguise since n even implies cos(πn) = and n odd implies cos(πn) =. converges. We can then apply the alternating series test to show the series Example 37. Determine whether the series converges or diverges ( ) n 2n 5n +. For part (i), we want b n+ b n 2(n + ) 5(n + ) + 2n 5n + (2n + 2)(5n + ) 2n(5n + 6) 0n 2 + 2n + 2 0n 2 + 2n 2 0. Hence, our original statement reduces (is equivalent to) a false statement, so the series fails part (i). 4
For the second part, note that, 2n 5n + = 2 5 0. Thus, this series fails both parts of the AST. (Note: if the series fails one part it fails the AST. I just wanted you to see an example where it fails both parts.) This alone does not guarantee divergence. However, the it 2n ( )n 5n + does not exist. Hence, by the Divergence Test, this series diverges. Example 38. Determine whether the series converges or diverges ( ( ) n sin2 π n) 2. n 2 In this case, condition (ii) holds (Squeeze Theorem). On the other hand, the series is not decreasing. Note that, if n is odd, then b n = n 2 series fails the Alternating Series Test. but if n is even then b n = 0. Thus, this However, this series still converges. Part of the reason is that this series is not really an alternating series. The negative terms are all 0, and so this is in fact a positive series. It is actually the series (2n ) 2 which converges by the Limit Comparison Test. Example 39. Determine the convergence of the series ( ) n+ n 2 n 3 +. Condition (ii) is easy to check and we omit that here. The real challenge is to check whether it is decreasing. It is enough, however, to show that the function f(x) = x2 x 3 + is decreasing (why?!). To determine intervals of increase and decrease, we look for critical points. Note that f (x) = x(2 x3 ) (x 3 +) 2. This has critical points at x =, 0, 3 2. We only care about x, so we check that f (x) < 0 for x > 3 2. Hence, f is decreasing for x 2. The statement of the theorem indicates that the decreasing condition must hold for all n, but as with most of our theorems, it is enough that it holds eventually. 5
Example 40. Test the series for convergence or divergence, ( n ) ( ) n + n. Let b n = n + n. We first check the it, ( n + n)( n + + n) n + n = n + + n = = 0. n + + n Now to show decreasing, we could argue as above. Instead, we will do this algebraically. The following statements are equivalent because all values considered are positive. b n+ b n n + 2 n + n + n n + 2 + n 2 n + (n + ) + 2 n n + 2 4(n + ) 2 n n + 2 2(n + ) n n + 2 (n + ) n(n + 2) (n + ) 2 0. Remark. Alternating series are particularly nice because there is an easy formula to determine a bound for the error of the Nth partial sum S N. Theorem 4 (Alternating series estimation theorem). If S = ( )n b n is the sum of a series satisfying the AST, then R N = S S N b N+. Remark. Another way of stating the above is that S N b N+ S S N b N. Example 42. Find the sum of the series ( ) n correct to three decimal places. n! First check the Alternating Series Test. The fact that the summation starts at n = 0 does not affect the AST. Set b n =. It is clear that n! b n = n! = 0. 6
Then b n+ b n (n + )! n! (n + )! n! n + 0 n. This last statement is true by assumption. Hence, {b n } is a decreasing sequence and so the series converges by the Alternating Series Test. Let S be the sum of the series. The idea of the above theorem is that we must find a b n such that the difference does not affect the third decimal place of the sum. This is a bit of trial and error. We compute the b n and find that b 7 = 0.0002. Hence, S S 6 b 7 = 0.0002. We compute S 6 0.368056. Hence, 0.367856 S 6 0.0002 S S 6 + 0.002 0.368256. Round both sides to three decimal places gives 0.368 S 0.368, so S 0.368. 6. Absolute convergence, ratio and root tests Remark. In the previous section, we ited ourselves to only series with positive terms. We now consider series with negative terms. Definition 8. A series a n is called absolutely convergent if the series of absolute values an is convergent. We say a n is conditionally convergent if a n converges but a n diverges. Example 43. The series ( ) n n 2 the alternating series test). is absolutely convergent. It is also convergent (by The alternating harmonic series converges but it is not absolutely convergent. Theorem 44. If a series a n is absolutely convergent, then it is convergent. 7
Proof. We know a n a n a n so 0 a n + a n 2 a n. Since a n converges then 2 an, so (a n + a n ) converges by the Comparison Test. Since an = (a n + a n ) a n, is the difference of two convergent series, then a n converges. Example 45. Study the convergence of cos n n 2. We have cos n, so cos n n 2 n. 2 Hence, the series is absolutely convergent by the Comparison Test, and therefore convergent. Remark. We will now consider two tests, the Ratio Test and the Root test, which can in certain cases determine whether a series is absolutely convergent. Theorem 46 (The Ratio Test). Assume the following it exists: L = a n+ a n. If L <, then the series a n is absolutely convergent. If L > or L =, then the series a n is divergent. If L =, then the Ratio Test is inconclusive. Example 47. Study the convergence of each example. n () 2 n Set a n = n. Then 2 n L = a n+ a n = (n + )/2 n+ n/2 n = (n + )2 n n2 n+ = 2. Since L <, then the series a n converges absolutely by the Ratio Test. (2) n 3 Set a n =. Now we compute n 3 L = a n+ a n = /(n + ) 3 /n 3 = n 3 (n + ) 3 =. 8
Thus, the Ratio Test is inconclusive in this case. Thankfully, since the terms are always positive, we know already that this series converges absolutely by the p-test. n! (3) 6 n Set a n = n!. Then 6 n L = a n+ a n = (n + )!/6 n+ n!/6 n = (n + )!6 n n!6 n+ = Since L =, then the series a n diverges by the Ratio Test. ( 3) n (4) n Set a n = ( 3)n n. Then L = a n+ a n = ( 3) n / n + ( 3) n / n = Since L >, then the series a n diverges by the Ratio Test. Remark. A quick reminder that for any finite number p, (np ) /n = n p/n =. n + 6 n3 n n + 3 n = 3. =. This can be shown by writing a function which models the sequence and L Hospital s Rule. Theorem 48 (The Root Test). Assume the following it exists: L = n a n. If L <, then the series a n is converges absolutely. If L > or L =, then the series a n diverges. If L =, then the Root Test is inconclusive. Example 49. Study the convergence of each example. ( ) n 2 n + (). 2n 2 + ( n Set a n = 2 + n. 2n +) Then 2 L = ( n n n2 + a n = 2n 2 + 9 ) n = n 2 + 2n 2 + = 2.
Since L <, then the series a n converges absolutely by the Root Test. ( 5) n (2). n 2 This one could be done (fairly easily) with the Ratio Test. But we ll use the Root Test along with the fact mentioned above. Set a n = ( 5)n. Then n 2 L = n n a n = ( 5) n = n 2 Since L >, then the series a n diverges by the Root Test. 5 = 5. n2/n 8. Power Series Remark. Recall that in a geometric series, the coefficients were assumed to be constant. We now generalize this and allow the coefficients to vary. We then have what is known as a power series. We will study when power series converge using many of the tools developed in previous sections. Definition 9. A power series with center a is a series of the form F (x) = c n (x a) n = c 0 + c (x a) + c 2 (x a) 2 +. The c n are constants called the coefficients of the series. Remark. Notice that, in contrast to our previous convention, we typically start a power series at the index n = 0. Example 50. You have already seen an example of a power series, the geometric series, cr n = cr n. Hence, c n = c for all n. We already know when this series converges, that is, when r <. In terms of power series, we call the radius of convergence and (, ) the interval of convergence. Remark. As we saw in the last example, a power series F (x) may converge for some values of x and diverge for other values of x. It turns out that there is always interval I, called the interval of convergence, such that F (x) converges for values of x in I and diverges for 20
values outside (there is a question of what happens at the endpoints which must be checked separately). Theorem 5 (Radius of covergence). Every power series F (x) has a radius of convergence R, which is either a nonnegative number of infinity. If R is finite, F (x) converges absolutely when x c < R and diverges when x c > R. If R =, then F (x) converges absolutely for all x. Remark. In each example, we must first find the radius of convergence. We then must check the endpoints individually. Example 52. Find the radius and interval of convergence for the series ( ) n x n n +. The center here is 0. We apply the Ratio Test, ( ) n+ x n+ /(n + 2) ( ) n x n /(n + ) = x n+ /(n + 2) x n /(n + ) = xn + n + 2 n + = x n + 2 = x. This power series converges when x < and diverges when x >, so the radius of convergence is. Now we need to check the endpoints x = and x =. When x =, this series is just the alternating harmonic series. On the other hand, when x =, this is the harmonic series. Hence, the interval of convergence is (, ]. Example 53. Find the radius and interval of convergence for the series n!x n. Again we apply the Ratio Test (this will be our primary tool, but occasionally we may apply the Root Test as well). (n + )!x n+ n!x n = (n + )x = x (n + ). This it is infinite unless x = 0. Thus, the radius of convergence of the series is 0. Example 54. Find the radius and interval of convergence for the series n (x + 2)n ( ). n2 n 2
Note that this series is forced to start at n =. However, that will not affect the radius or interval of convergence. We will apply the Root Test here, but the Ratio Test would also work. n (x + 2)n ( )n n2 n = x + 2 n n2 = x + 2 2 n n = Thus, we need x+2 2 < which is equivalent to x + 2 < 2. Thus, R = 2. x + 2. 2 The condition x + 2 < 2 implies 2 < x + 2 < 2 or 4 < x < 0. Thus, we need to check the series when x = 4 and when x = 0. When x = 0 this is the alternating harmonic series. When x = 4 it is the harmonic series. Thus the interval is ( 4, 0]. Example 55. Find the radius and interval of convergence for the series ( ) n x2n (2n)!. Applying the ratio test, ( ) n+ x 2(n+) /(2(n + ))! ( ) n x 2n /(2n)! = x 2(n+) (2n)! x 2n (2(n + ))! = x 2 (2n + 2)(2n + ) = x 2 (2n + 2)(2n + ) = 0. Hence, the series converges for all x and the radius of convergence is R =. Example 56. Find the radius and interval of convergence for the series n 5 (x 2) n. 5 n Again applying the Ratio Test, (n + ) 5 (x 2) n+ /5 n+ n 5 (x 2) n /5 n Hence, R = 5 and the endpoints are x = 3, 7. = x 2 5 (n + ) 5 = n 5 x 2. 5 When x = 7, the series becomes n 5 and when x = 3, the series becomes ( ) n n 5. Both diverge by the Divergence Test. Thus, the interval of convergence is ( 3, 7). 22
9. Representations of functions as power series Remark. Suppose x <. Then x = x n = + x + x 2 + x 3 +. Said another way, the function /( x) may be represented by the power series. This has many practical applications. In particular, power series are often better to work with because they resemble polynomials. Hence, if a function like e x can be represented by a power series (and it can) then we can approximate e 5 by using the power series representation. In this section we will study multiple techniques for representing certain functions by power series. We expand this in the next section using Taylor series. Example 57. Express /( + x 2 ) as the sum of a power series and find its interval of convergence. We apply substitution to the power series form of /( x) by replacing x with x 2. + x = 2 ( x 2 ) = ( x 2 ) n = ( ) n x 2n = x 2 + x 4 x 6 + x 8 +. Because this is a geometric series, it converges when x 2 <, that is, when x 2 <. But this is just equivalent to x <. So the interval of convergence is (, ). Example 58. Express /(x 3) as the sum of a power series and find its interval of convergence. We first need to write this in the form of the sum of a geometric series. We do this by factoring out a 3 from the denominator. ( ) x 3 = ( 3)( (x/3)) = 3 x = 3 3 ( x ) n = 3 3 3 n xn. This series converges when x/3 < or x < 3. Thus, the interval of convergence is ( 3, 3). Example 59. Express x 2 /(x 3) as the sum of a power series and find its interval of convergence. We can use our work from the previous example. ( ) x 2 x 3 = x2 x 3 = x2 3 3 n xn = 3 The interval of convergence is ( 3, 3). 23 3 n xn+2 = 3 n=2 3 n 2 xn.
Theorem 60 (Term-by-term differentiation and integration). If the power series F (x) = c n (x a) n has radius of convergence R > 0, then F (x) is differentiable (and therefore continuous) on (a R, a + R) (or all x if R = ). Furthermore, we can integrate and differentiate term by term. For x (a R, a + R), () F (x) = c + 2c 2 (x a) + 3c 3 (x a) 2 + = (2) nc n (x a) n (x a) 2 F (x) dx = C + c 0 (x a) + c + = C + 2 Moreover, these series have the same radius of convergence R. as a power series by differentiating the power series represen- Example 6. Express ( x) 2 tation for. x Using term-by-term differentiation, ( x) = d 2 dx c n (x a)n+ n + ( x ) = d dx ( + x + x2 + x 3 + ) = 0 + + 2x + 3x 2 + = nx n. Since term-by-term differentiation does not affect the radius of convergence, then we get radius of convergence R =. Remark. The radius of convergence will remain the same when we differentiate and integrate, but the interval of convergence may change. Check endpoints! Example 62. Find a power series expansion for arctan(x). By Example 57, ( + x 2 ) = x 2 + x 4 x 6 + when x <. Then arctan x = ( + x 2 ) dx = ( x 2 + x 4 x 6 + ) dx = C + x x3 3 + x5 5 x7 7 +. When x = 0, C = arctan(0) = 0. Hence, ( ) n x 2n+ arctan x =. 2n + This expansion is valid for < x < but not at the endpoints. 24
Example 63. Find a power series expansion for ln( + x). We have + x = ( x) = ( ) n x n and this series had radius of convergence. Now ln( + x) = + x dx = ( x + x 2 x 3 + x 4 + ) dx ) = (x x2 2 + x3 3 x4 4 + x5 5 + + C = C + ( ) n xn+ n +. When x = 0, ln( + x) = ln() = 0. Hence, C = 0. This has radius of convergence R =. 0. Taylor Series Remark. For a function f(x), the linearization at x = a is the function L(x) = f (x)(x a) + f(a). The linearization approximates the function near x = a. But more than that, f(x) and L(x) agree at x = a and their derivatives agree at x = a. That is, f(a) = L(a) and f (a) = L (a). The question in this section is whether we can construct functions that agree at higher derivatives as well and how well do these coincide with each other. It turns out that functions that can be represented by a power series can always be represented by a very special series. Remark. Suppose f can be represented by a power series centered at a, that is f(x) = c n (x a) n, x a < R. Can we determine a formula for the c n in general? First note that f(a) = c 0. Now if we take the derivative we find, f (x) = nc n (x a) n. Thus, f (a) = c. Similarly, f (x) = n(n )c n (x a) n 2, n=2 25
so f (a) = 2c 2. Continuing in this way, we find that f (n) (a) = n!c n. Said another way, c n = f (n) (a) n!. Theorem 64 (Taylor s Formula). If f has a power series representation centered at a, that is, f(x) = c n (x a) n, x a < R, then its coefficients are given by the formula, c n = f (n) (a) (x a) n. n! Remark. The above series is called the Taylor Series of f at a and, when a = 0, it is called the Maclaurin series. Example 65. Find the Maclaurin Series of f(x) = e x and its radius of convergence. First note that f (n) (x) = e x for all n and so f (n) (0) = for all n. Hence, e x x n = n!. Using the ratio test, we find the radius of convergence is R = (this was a WebAssign problem). Example 66. Find the Maclaurin series for sin x. Let f(x) = sin x, then f (x) = cos x, f (x) = sin x, f (x) = cos x and f (4) = sin x. Hence, this cycle repeats indefinitely. We have f(0) = 0, f (0) =, f (0) = 0, and f (0) =. Thus, the Maclaurin series for sin x is ( ) n x 2n+ (2n + )!. Example 67. Find the Taylor series of f(x) = /x at a = 3. We have f(x) = x, f (x) = x 2, f (x) = 2x 3, f (x) = 6x 4. Thus, we conclude that f (n) (x) = ( ) n (n!)x (n+). Hence, f (n) ( 3) = ( ) n (n!)( 3) (n+) = (n!)3 (n+). Therefore, the Taylor series expansion of f(x) = /x at a = 3 is (n!)3 (n+) (x ( 3)) n = n! 3 (x + n+ 3)n. 26
Remark. Our understanding of Taylor series gives us a way to realize a very important series which you are probably already familiar with. Example 68 (The Binomial Series). Find the Maclaurin series for f(x) = ( + x) k, where k is any real number. so that We can show that Hence, the Maclaurin series of f(x) is f (0) (x) x n = n! f (n) (x) = k(k )(k 2) (k n + )( + x) k n, f (0) (x) = k(k )(k 2) (k n + ). k(k )(k 2) (k n + ) x n. n! The coefficients in this series are known as binomial coefficients, ( ) k k(k )(k 2) (k n + ) =. n n! Hence, we can write the series in shorthand as, ( ) k ( + x) k = x n. n Remark. Now we turn to the question of whether a function is truly represented by its Taylor/Maclaurin series. One way to do this is to find the radius of convergence. In the first two examples, the radius of convergence is R =. On the other hand, in the previous example, the radius of convergence is 3. There is a shortcut which can save time in certain instances. Remark. We call the partial sums of the Taylor series the Taylor polynomials of f at a. Let T n (x) represent the nth Taylor polynomial of a function f. Then R n (x) = f(x) T n (x) is called the remainder of the Taylor series. Theorem 69. If f(x) = T n (x) + R n (x) and R n(x) = 0 for x a < d, then f is equal to the sum of its Taylor series on the interval x a < d. Theorem 70. Suppose there exists M > 0 such that f (n+) (x) M for all n and all x such that x a d, then f(x) is equal to the sum of its Taylor series on (a d, a + d). 27
Remark. This actually follows from two facts: The first is Taylor s Inequality which states that and from the fact that for any real number x. R n (x) M (n + )! x a n+, xn /n! = 0 Example 7. Prove e x equals the sum of its Taylor Series at x = a. We have f (k) (x) e a+r for x (a R, a + R). Remark. One way of finding Taylor/Maclaurin series is to obtain them from old ones. In particular, one could substitute/integrate/derive known Taylor series to obtain the Taylor series for a new function. Example 72. Show that sin x equals the sum of its Maclaurin series at a = 0 and, find the Maclaurin series for x sin x. Let g(x) = sin x. From above, we know that the Maclaurin series for sin x is sin x = ( ) n x 2n+ (2n + )!. Since g (k) (x) for all x, then sin x equals the sum of its Maclaurin series. Hence, x sin x = ( ) n x2(n+) (2n + )!. Example 73. Evaluate the following indefinite integral as an infinite series. e x2 dx We have Hence, by substitution, e x = e x2 = Now, by term-by-term integration, e x2 dx = C + x n n!. ( ) n x2n n!. ( ) n x 2n+ n!(2n + ). 28