Chapter 8. Infinite Series

Transcription

1 8.4 Series of Nonnegative Terms Chapter 8. Infinite Series 8.4 Series of Nonnegative Terms Note. Given a series we have two questions:. Does the series converge? 2. If it converges, what is its sum? Corollary of Theorem 5. Aseries of nonnegative terms converges if its partial sums are bounded from above. Proof. Theorem 5 (of section 8.2) implies that a monotonic increasing sequence which is bounded above must converge. A positive term series will have partial sums which form a monotonic increasing sequence. Since we have hypothesized that the sequence of partial sums is bounded, the result follows. Q.E.D.

2 8.4 Series of Nonnegative Terms 2 Theorem. The Integral Test Let { } be a sequence of positive terms. Suppose that = f(n), where f is a continuous, positive, decreasing function of x for all x N (N a positive integer). Then the series and the integral f(x) dx both converge or both diverge. n=n Proof. Since a finite number of terms does not affect the convergence of a series, we may assume that N = without loss of generality. Under the hypotheses of f as continuous and decreasing, we can consider the following rectangles: N Figure 8.3a page 64 The areas of the rectangles are a,a 2,a 3,...,, and since f is decreasing, these rectangles are circumscribed over f and we have n+ f(x) dx a + a

3 8.4 Series of Nonnegative Terms 3 If we consider inscribed rectangles, then we have: Figure 8.3bpage 64 Excluding a, we see that a 2 + a 3 + a or that Therefore we know that If n+ n a + a 2 + a a + f(x) dx, n f(x) dx. f(x) dx a + a 2 + a a + n f(x) x. f(x) dx is finite, then the right-hand inequality shows that is finite. If f(x) dx is infinite, then the left-hand inequality shows that is infinite. Q.E.D.

4 8.4 Series of Nonnegative Terms 4 Example. Number 4 page 649. Theorem. p-series A series of the form n p = p + 2 p + 3 p + + n p + is called a p-series. Ap series converges if p> and diverges if p. Proof. We prove this using the Integral Test. First, suppose p.then ( dx b ) ( ) x = lim dx x p+ b = lim p b x p b p + ( ) ( ) = lim b p (b p+ ) = lim b p b p p =, p >, p <. Therefore both the integral and the series converge if p>, and both diverge if p<. Next, suppose that p =. Then dx x = lim b b dx x = lim ( ) ln x b = lim (ln b ln ) =. b b By the integral test, the series diverges when p =. Q.E.D. Definition. The p-series with p =istheharmonic series.

5 8.4 Series of Nonnegative Terms 5 Note. Let s briefly explore the rate at which the harmonic series diverges. Example 3 on page 642 asks how many terms must we add in the harmonic series to get a partial sum greater than 20. Consider these two graphs: Figure 8.4 page 642. We see that the 4th partial sum is less than + ln 4, and in general the n th partial sum will be less than + ln n. Therefore we need at least +lnn>20, or n>e 9 78, 482, 30. We can use a similar argument with circumscribed rectangles to see that the n th partial sum is greater than ln(n + ), and so we find that to get the partial sum greater than 20, we would need at most n = e , 65, 94.

6 8.4 Series of Nonnegative Terms 6 Theorem. Direct Comparison Test Let be a series with no negative terms. (a) converges if there is a convergent series c n with c n for all n>n, for some integer N. (b) diverges if there is a divergent series of nonnegative terms d n with d n for all n>n, for some integer N. Proof. For part (a), the partial sums of are bounded above by M = a + a c n. n=n+ Therefore by the corollary to Theorem 5, the result holds. For part (b), the partial sums of are not bounded above (for if they were, then the partial sums of d n would be bounded and it would be convergent). Therefore diverges. Q.E.D. Example. Number 4 page 649.

7 8.4 Series of Nonnegative Terms 7 Theorem. Limit Comparison Test Suppose that > 0andb n > 0 for all n N (N a positive integer).. If lim = c, 0<c<, then and b n both converge or n b n both diverge. 2. If lim =0and b n converges, then converges. n b n 3. If lim = and n b n b n diverges, then diverges. Proofof().Since c/2 > 0, there exists an integer N such that for all n>n we have c b n < c ɛ. So for n>n it follows that 2 c 2 < c< c b n 2, c 2 < < 3c b n 2, ( ( ) c 3c b n < < b n. 2) 2 ( ) 3c If b n converges then b n converges and 2 the Direct Comparison Test. If b n diverges, then and diverges by the Direct Comparison Test. converges by ( c b n diverges 2) Q.E.D.

8 8.4 Series of Nonnegative Terms 8 Example. Number 6 page 649. Theorem. The Ratio Test Let be a series with positive terms and suppose that Then. the series converges if ρ<, + lim = ρ. n 2. the series diverges if ρ>orifρ is infinite, and 3. the test is inconclusive if ρ = (that is, the series could diverge or converge the Ratio Test tells us nothing). Proof. (a) Let r be umber between ρ and. Then the number + ɛ = r ρ is positive. Since lim = ρ, then there exists positive n integer N such that for all n N we have + / within ɛ of ρ. In particular, for n N we have + <ρ+ ɛ = r. Thatis, a N+ < ra N a N+2 < ra N+ <r 2 a N a N+3 < ra N+2 <r 3 a N

9 8.4 Series of Nonnegative Terms 9. a N+m < ra N+m <r m a N. If we define the series c n a + a a N + a N ( + r + r 2 + ), then we see that this new series converges, for it is (eventually) a geometric series with ratio r between 0 and. Therefore by the Direct Comparison Test, the series converges. (b) With <ρ, we must eventually have (that is, for all n M where M is some positive integer) + >. That is, 0 <a M <a M+ < a M+2 <. Therefore the sequence { } either diverges or has a limit greater than 0. So by the Test for Divergence, the series diverges. (c) Consider the two series n and n2. For both series, ρ =, but the first series diverges, while the second series converges. Q.E.D.

10 8.4 Series of Nonnegative Terms 0 Note. The Ratio Test (if applicable) is easier to use than the Direct Comparison Test. This is because you don t need to find a second series which has the appropriate behavior (in terms of convergence or divergence) and satisfies the appropriate inequalities. Example. Number 24 page 649. Theorem. The n th -Root Test Let be a series with 0forn N and suppose that lim n = n ρ. Then (a) the series converges if ρ<, (b) the series diverges if ρ>orρ is infinite, and (c) the test is inconclusive if ρ =. Note. Again, the Root Test doesn t require a second series and is easier to use than the Direct Comparison Test. Example. Number 34 page 649.