Sequences and Series

CHAPTER Sequences and Series.. Convergence of Sequences.. Sequences Definition. Suppose that fa n g n= is a sequence. We say that lim a n = L; if for every ">0 there is an N>0 so that whenever n>n;ja n Lj <". If lim a n = L; we say that the sequences converges, otherwise it diverges. Example. Use the deönition of convergence to show that lim n =0 Example. Use the deönition of convergence to show that the sequence f( ) n g n= does not converge. Example 3. Use the deönition of convergence to show that lim n p 3=: Theorem. If the sequence fa n g n= converges, then there is exactly one number L so that lim a n = L:.. Special Limits. Theorem. If x>0; then lim n p x =: Theorem 3. If r>0 is a real number, then lim n r =0 Theorem 4. If jxj < ; then lim x n =0. Theorem 5. If x is any real number, then lim x n =0:.3. Computing with Limits. Definition. A sequence fa n g n= is bounded from above, if there is a number B so that a n B for all n : A sequence fa n g n= is bounded from below, if there is a number B so that a n B for all n : If there is a number B 0 so that B a n B (or, equivalently, ja n j B), then the sequence fa n g n= is called a bounded sequence. Example 4. The sequence fng n= above, and not bounded. is bounded from below, not bounded from Example 5. The sequence n n= is bounded. Theorem 6. Every convergent sequence is bounded. v

vi. SEQUENCES AND SERIES Theorem 7. Suppose that lim a n = L and lim b n = M: Further, suppose that k is a constant. Then lim n = k lim n = kl lim n + b n ) = lim n + lim n = L + M lim n b n ) = lim n lim n = L M lim nb n = lim n lim n = LM lim nj = lim n = jlj Furthermore, if b n 6=0for all n; and if M = lim b n 6=0; then a n lim = lim a n = L b n lim b n M Sometimes sequences a numbered not starting with n =; but with a larger value of n : Example 6. Show that the sequence n o converges and Önd the limit. Justify each step carefully! 3n +7n+44 n 65n+93 n=3 Example 7. The example of a n = ( ) n and b n = ( ) n+ shows that lim (a n + b n ) might exist, while the two limits lim a n and lim b n do not exist. So one has to be careful how one reads the equation lim (a n + b n )= lim a n + lim b n : The correct way to read them is: If the limits lim a n and lim b n both exist, then lim (a n + b n ) also exists, and lim (a n + b n )= lim a n + lim b n : Theorem 8. Suppose that lim a n and lim b n both exist. If a n b n for all values of n; then lim a n lim b n.4. Criteria for Convergence. Theorem 9 (Squeeze Theorem). Suppose that a n b n c n for all n N for some N: Moreover, suppose that lim a n and lim c n both exist and have a common value L: Then the sequence fb n g n= also converges, and lim b n = L: Corollary. If fa n g n= is a sequence, then lim ja n j =0if and only if lim a n =0: Example 8 (Babylonian Roots). The Babylonians used the following method to approximate square roots: Let x 0 be any non-negative real number, and let w 0 > 0 be arbitrary. We deöne recursively w n+ = w n + xwn Then the inequality a + b p ab; which holds for all positive real numbers a and b; implies that w n+ = r w n + xwn x w n = p x w n

. SEQUENCES vii From this inequality we conclude that w n+ p x for n 0 w n p x for n We use this inequality to show by induction that w n p x + w for n n Indeed, for n =; we have w p x + w = p x + w : If we already know that 0 w n p x + w ; then also n w n + xwn w n+ = px w + n + x w n px w + n + p x xx = p x + w n So, we have p x wn p x + w n The squeeze theorem now implies that lim w n = p x Definition 3. (i) A set A of real numbers has an upper bound B; if x B for all x A: (ii) An upper bound B 0 of A is called a least upper bound of A; if B 0 is an upper bound, and B 0 B for any other upper bound B of A: Example 9. Find upper bounds and least upper bounds of the following sets: (i) A = + n : n (ii) A = ( ) n n : n The real number system is characterized by the fact that every nonempty subset A that has an upper bound also has a least upper bound. This is something that cannot be proven, rather it is part of the deönition of the real number system. So we will call it an axiom: AXIOM (Least upper bound axiom) Every nonempty set of real numbers that has an upper bound has a least upper bound. Theorem 0. If B is the least upper bound of the set A and if ">0; then there is at least one number x A so that B "<x<b: Related to upper bounds are lower bounds:

viii. SEQUENCES AND SERIES Definition 4. (i) A set A of real numbers has an lower bound C; if C x for all x A: (ii) An upper bound C 0 of A is called a greatest lower bound of A; if C 0 is an lower bound, and C C 0 for any other lower bound C of A: Example 0. Find lower bounds and greatest lower bounds of the following sets: (i) A = + n : n (ii) A = ( ) n n : n The following statement is NOT an axiom, but it can be shown from the least upper bound axiom: Theorem. Every nonempty set of real numbers that has a lower bound has a greatest lower bound. Definition 5. A sequence fa n g n= is called increasing, if a n a n+ for all n: Similarly, the sequence fa n g n= is called decreasing if a n a n+ for all n: A sequence that is either increasing or decreasing is called monotonic. Example. The sequence n is increasing, and the sequence n+ n n= n is decreasing. n= Theorem. If fa n g n= is increasing and bounded from above, then the sequence converges. Similarly, if fa n g n= is decreasing and bounded from below, then the sequence also converges. Example. Show that n =n n= converges. Example 3. Show that f=n n g n= converges. Example 4 (Eulerís Number). Let e n = 0! +! + + Prove that fe n g n= converges (Hint: Show that fe ng n= is increasing and bounded from above by 3:).5. Sequences and Continuous Functions. Review the " - deönition of continuous functions. Theorem 3. Let f (x) be a function that is continuous at x 0 : If fa n g n= is a sequence in the domain of f with lim a n = x 0 ; then lim f (a n ) exists and lim f (a n )=f (x 0 ) : Example 5. Show that lim sin n exists and is equal to 0: q n n Example 6. Show that lim exists and Önd the limit. n +n+

. SERIES ix. Series.. DeÖnition of Convergence of Series. If we talk about series, we start again with a sequence: Definition 6. Let fa n g be a sequence. Then the kth partial sum of fa k g k= is the Önite sum kx s k = We say that the series P a n converges, if lim k! s k exists. In this case, a n a n = lim s k = lim k! k! If the limit lim s n does not exist, then we say that the series P a n diverges. kx a n to x Definition 7. The series P xn is called the geometric series. Theorem 4. The geometric series P xn =+x + x + x 3 + ::: converges for jxj < and diverges for all other values of x: Example 7. Show that P n n Example 8. Show that P n= n +n diverges. converges to : Theorem 5. If P a n converges, then lim a n =0 The converse of the previous theorem is not true: There are series P a n for which lim a n =0; yet nevertheless the series P a n diverges. The most well-known such series is the harmonic series: Definition 8. The series P n= n =+ + 3 + 4 + ::: is called the harmonic series. Theorem 6. The harmonic series diverges. Similarly to the rules for Önding limits of sequences, there are rules for Önding values of inönite series: Theorem 7. Let P a n and P b n be two convergent series, and let c be a constant. Then (i) the series P ca n is also convergent, and P ca n = c P a n; and (ii) the series P (a n + b n ) is also convergent, and P (a n + b n )= P P a n+ b n:.. Series with Nonnegative Terms.

x. SEQUENCES AND SERIES... Basic Test. Theorem 8. A series with nonnegative terms a n 0 converges if and only if the sequence of partial sums is bounded. As an application of the previous theorem, we show Theorem 9. The series P converges. Of course, this does not tell us what the limit might be. Later, we show that P = e:... The Integral Test. Theorem 0 (Integral Test). If f is continuous, decreasing and positive on [; ) ; then P n= f (n) converges if and only if R f (x) dx converges. The integral test allows another proof of the fact that the harmonic series diverges: Example 9. The harmonic series P n= n diverges. Definition 9. If p 0 is a real number, then P n= n p Theorem. If p 0; then the p-series P p>: n= n p is called a p-series. converges if and only if Example 0. Show that P n= 5 n 4 converges and that P n= p n diverges...3. Comparison Tests. Theorem. Let fa n g n= and fb ng n= be two nonnegative sequences so that 0 a n b n : () If the series P n= b n converges, then so does P n= a n. () If the series P n= a n diverges, then so does P n= b n. Theorem 3. Let fa n g n= and fb nng n= o be two nonnegative sequences so that a for a certain value of N the sequence n bn is bounded. n=n () If the series P n= b n converges, then so does P n= a n. () If the series P n= a n diverges, then so does P n= b n. Theorem 4. Let fa n g n= and fb ng n= be two nonnegative sequences so that a lim n bn < : () If the series P n= b n converges, then so does P n= a n. () If the series P n= a n diverges, then so does P n= b n. Example. Does P n= n ln n converge? Example. Does P jsin nj n= n converge? Example 3. Does P p n= converge? n 3 Example 4. Does P p n= converge? n +3

. SERIES xi.3. Alternating Series. Definition 0. Let fa n g be a sequence of nonnegative numbers. Then each of the series P ( )n a n and P ( )n+ a n is called an alternating series. Theorem 5 (Leibniz Test). Suppose that fa n g is a sequence so that 0 a P n+ a n for all values of n 0 and so that lim a n =0: Then the series ( )n a n converges. Moreover, for each value of N; NX ( ) n a n ( ) n a n a N+ Example 5. The alternating harmonic series P n= ( )n+ a n converges. Example 6. Approximate the alternating harmonic series to one decimal place..4. Absolute Convergence. Definition. Let fa n g be a sequence of number. If P ja nj converges, then we say that P a n converges absolutely. If P P a n converges, but ja nj does not, then we say that P a n converges conditionally. Example 7. The series P n= sin n n converges absolutely. Example 8. The alternating harmonic series converges conditionally. The next theorem seems obvious, but is not: Theorem 6. Every absolutely converging series converges..5. The Ratio and Root Tests. a Theorem 7 (The Ratio Test). Suppose that lim n+ a n = L: () If L<; the series P a n converges absolutely. () If L>; the series P a n diverges. (3) If L =; this test gives no information. Theorem 8. The series P xn converges absolutely for each value of x: Theorem 9. Suppose that lim ja n j =n = L: () If L<; the series P a n converges absolutely. () If L>; the series P a n diverges. (3) If L =; this test gives no information. Example 9. Analyze P xn n n :.6. Power Series. Definition. A power series is a series of the form a n x n with the understanding that a n may depend on n but not on x: Example 30. P converge? xn n is a power series. For which value of x does this series

xii. SEQUENCES AND SERIES Example 3. The series P ( )n x n+ (n+)! is a power series. Find the values of the a n ís. Example 3. The series P xn is a power series. Find an explicit formula for f (x) : Theorem 30. If the power series P a nx n converges for x = x 0 ; then it converges absolutely for all values of x with jxj < jx 0 j : And, if P a nx n diverges for x = x ; then it diverges for all values of x with jxj > jx j : Theorem 3. If P a nx n is a power series, then there is a number R with 0 R so that the series converges absolutely for jxj <Rand diverges for jxj >R: Definition 3. The number R from the previous theorem is called the radius of convergence of the power series P a nx n : The interval of convergence consists of all values of x for which P a nx n converges. So each power series has a radius of convergence R: For x = R; the power series might converge absolutely, conditionally or it might diverge. So the interval of convergence could be either of the following intervals: ( R; R) ; [ R; R] ; ( R; R] or [R; R): Example 33. Find the interval of convergence of the power series P n= ( )n x n n : Theorem 3. Let P a nx n be a power series so that R = lim a n a n+ exists (including the value + for R). Then R is the radius of convergence. Theorem 33. Let P a nx n be a power series so that R = lim ja nj =n exists (including the value + for R). Then R is the radius of convergence. Example 34. Find the radius of convergence and the interval of convergence of P nxn : Example 35. Find the radius of convergence and the interval of convergence of P n x n : Example 36. Find the radius of convergence and the interval of convergence of P n x n : Some power series are not "centered at 0": Definition 4. A power series centered at a is a series of the form a n (x a) n with the understanding that a n may depend on n but not on x: So ordinary power series are power series that are centered at 0: Like ordinary power series, power series centered at a also have a radius of convergence and an interval of convergence.

. SERIES xiii of Example 37. Find the radius of convergence and the interval of convergence Also, Önd an explicit formula for f (x) : (x + ) n 3 n+.7. Calculus with Power Series. The proofs of the following theorems SHOULD NOT BE SKIPPED! Theorem 34. Let a n (x a) n be a power series with radius of convergence R: Then f (x) is a continuous function for jx aj <R: The following two theorem say that power series can be di erentiated and integrated like polynomials: Theorem 35. Let a n (x a) n be a power series with radius of convergence R: Then g (x) = na n (x a) n has also R as radius of convergence. Moreover, f (x) is di erentiable for jx aj <R and f 0 (x) =g (x) : Theorem 36. Let a n (x a) n be a power series with radius of convergence R: Then a n F (x) = (x a)n+ n + has also R as radius of convergence. Moreover, F (x) is di erentiable for jx aj < R and F 0 (x) =f (x) : Example 38. Starting with the geometric series P xn = x ; we Önd that x n+ F (x) = n + is an antiderivative for x : Hence Since F (0) = 0; we Önd that F (x) = ln ( x)+c 0=F (0) = ln ( 0) + C = C

xiv. SEQUENCES AND SERIES Hence ln ( x) = x n+ n + This series converges for x < ; and therefore has [ ; ) as interval of convergence. It is tempting to write ( ) n+ ( ) n ln () = ln ( ( )) = = n + n but we have to realize that the endpoints of the interval of convergence are NOT covered by the previous theorem. It requires extra work to prove that indeed ( ) n ln () = n Theorem 37. We have ln ( x) = n= x n+ n + n= for <x< After all this preparation, we Önally come to the Örst "useful" application of series. The next example gives an inside why series and sequences have been originally been developed. Example 39. Find an approximation for ln 3 which is correct to two decimal places..8. Taylor Series. As we found out in the last example, power series can be used to compute function values. So, for a given function f (x) ; we would like to Önd a power series for f (x) : a n (x a) n and use this power series to compute values of f (x) : Theorem 38. If a n (x a) n has a radius of convergence R>0; then a n = f (n) (a) Definition 5. If f (x) is a function that has derivatives at x = a of arbitrary high order, and if a n = f (n) (a) ; then the series g (x) = a n (x a) n is called the Taylor series for f (x) at x = a: A Taylor series for f (x) at x =0is called the Maclaurin series for f (x) :

. SERIES xv IT IS NOT CLEAR THAT g (x) ; even in cases where g (x) has as radius of convergence! An example is the function exp x x 6= 0 0 x =0 y 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. -5-4 -3 - - 0 3 4 5 In this case, f (n) (0) exists for all values of n and So the Maclaurin series for f (x) is which is di erent from f (x) : a n = f (n) (0) g (x) = =0 0 x n =0 x Example 40. Find the Taylor series for x Example 4. Find the Maclaurin series for sin x: Example 4. Find the Maclaurin series for e x : centered at : At least in the last two examples, the Maclaurin series for sin x and e x may have nothing to do with sin x and e x ; as the example of the function exp =x taught us. We will further investigate this problem in the next section:

xvi. SEQUENCES AND SERIES.9. Taylorís Theorem. Theorem 39 (Taylor). Assume that f (x) is deöned on an interval I and that the (N + ) st derivative f (N+) (x) exists on I and is continuous. If x and a are two arbitrary points in I; then NX f (n) (a) (x a) n + R N where the remainder R n is given by either of the following two formulas: (i) or (ii) R N = f (N+) (z) (N + )! (x a) N+ for a value z between a and x Z x R N = (x t) N f (N+) (t) dt N! a In particular, f (x) can be represented by its Taylor series centered at a if and only if lim N! R n =0for all values of x: Theorem 40. We have the following representations: exp (x) = x n sin x = ( ) n x n+ (n + )! cos x = ( ) n x n (n)! Example 43. Compute an approximation of sin (0:) and Önd the error. Theorem 4. We have ln = ( ) n (Figure out that the remainder in Taylorís formula converges to 0 at a =:) n= Example 44. Compute an approximation of ln : Compute an approximation of ln 3: Compute an approximation of ln 6: etc Example 45. Find a Taylor series for x sin ( x) : Example 46. Find the Taylor series for cosh x and sinh x: Example 47. Find a Taylor series for (+x) :.0. Power series and Di erential Equations. Example 48. Use power series to solve y 0 = y n y (0) =

. SERIES xvii Example 49. Use power series to solve Do more examples like that! y 0 = xy y (0) =