Chapter 8 Infinite Series

Section 8.2 Convergence Tests

(Comparison Test) Theorem 8.2.1 Proof: In this section we derive several tests that are useful in determining convergence. (Comparison Test) Let  an and  bn be infinite series of nonnegative terms. That is, an  0 and bn  0 for all n. Then If  an converges and 0  bn  an for all n, then  bn converges. If  an = +  and 0  an  bn for all n, then  bn = + . Theorem 8.2.1 Proof: Since bn  0 for all n, the sequence (tn) of partial sums of  bn is an increasing sequence. In part (a) this sequence is bounded above by the sum of the series  an , so (tn) converges by the monotone convergence theorem (4.3.3). Thus  bn converges. In part (b) the sequence (tn) must be unbounded, for otherwise  an would have to converge. But then lim tn = +  by Theorem 4.3.8, so that  bn = + . 

Example 8.2.2 Definition 8.2.4 Consider the series . For all n  we have In Example 8.1.1 we saw that the series converges, so must also converge. Definition 8.2.4 If  | an | converges, then the series  an is said to converge absolutely (or to be absolutely convergent). If  an converges but  | an | diverges, then  an is said to converge conditionally (or be conditionally convergent).

Theorem 8.2.5 If a series converges absolutely, then it converges. Proof: Suppose that  an is absolutely convergent, so that  | an | converges. By the Cauchy criterion for series (Theorem 8.1.6), given any  > 0, there exists a natural number N such that n  m  N implies that | |  am | + … + |  an | | < . But then by the triangle inequality, so  an also converges.  When the terms of a series are nonnegative, convergence and absolute convergence are really the same thing. Thus the comparison test can be viewed as a test for absolute convergence. If we have a series  bn with some negative terms, the corresponding series | bn | will have only nonnegative terms, and we can use the comparison test on it.

If it happens that 0  |  bn  |  an for all n and if  an converges, then we can conclude that | bn | converges. That is,  bn converges absolutely. In fact, changing the first few terms in a series will affect the value of the sum of the series, but it will not change whether or not the series is convergent. So, given a nonnegative convergent series  an and a second series  bn, to conclude that  | bn | is convergent it suffices to show that 0  | bn |  an for all n greater than some N. By using the comparison test and the geometric series, we can derive two more useful tests for convergence.

(Ratio Test) Theorem 8.2.7 Proof: Let  an be a series of nonzero terms. If lim sup | an + 1/an | < 1, then the series converges absolutely. If lim inf | an + 1/an | > 1, then the series diverges. Otherwise, lim inf | an + 1/an |  1  lim sup | an + 1/an |, and the test gives no information about convergence or divergence. Theorem 8.2.7 Proof: (a) Let lim sup | an + 1/an | = L. If L < 1, then choose r so that L < r < 1. By Theorem 4.4.11(a) there exists N  such that n  N implies that | an + 1/an |  r. (If there were infinitely many terms | an + 1/an | greater than r, then a subsequence would converge to something greater than or equal to r, a contradiction to r being greater than the lim sup | an + 1/an |.) That is, n  N implies | an + 1 |  r |an |. So, | aN + 1 |  r |aN |, and | aN + 2 |  r |aN + 1 |  r2|aN |, etc. It follows easily by induction that | aN + k |  rk |aN | for all k  . Since 0 < r < 1, the geometric series   rk is convergent. Thus  | aN | rk is also convergent, and  an is absolutely convergent by the comparison test.

Theorem 8.2.7 (Ratio Test) Proof: Let  an be a series of nonzero terms. If lim sup | an + 1/an | < 1, then the series converges absolutely. If lim inf | an + 1/an | > 1, then the series diverges. Otherwise, lim inf | an + 1/an |  1  lim sup | an + 1/an |, and the test gives no information about convergence or divergence. Proof: (b) If lim inf | an + 1/an | > 1, then it follows that | an + 1| > | an | for all n sufficiently large. Thus the sequence (an) cannot converge to zero, and by Theorem 8.1.5 the series  an must diverge. (c) This follows from the observation that the series  1/n2 converges and the harmonic series  1/n diverges. In both series we have lim | an + 1/an | = 1. 

(Root Test) Theorem 8.2.8 Proof: Given a series  an, let  = lim sup | an |1/n. If  < 1, then the series converges absolutely. If  > 1, then the series diverges. Otherwise,  = 1, and the test gives no information about convergence or divergence. Theorem 8.2.8 Proof: (a) If  < 1, choose r so that  < r < 1. By Theorem 4.4.11(a) we have |  an |1/n  r for all n greater than some N. That is, |  an |  r n for all n > N. Since 0 < r < 1, the geometric series  r n is convergent, so  an is absolutely convergent by the comparison test. (b) If  > 1, then |  an |1/n  1 for infinitely many indices n. That is, |  an |  1 for infinitely many terms. Thus the sequence (an) cannot converge to zero and, by Theorem 8.1.5, the series  an must diverge.

Theorem 8.2.8 (Root Test) Proof: Given a series  an, let  = lim sup | an |1/n. If  < 1, then the series converges absolutely. If  > 1, then the series diverges. Otherwise,  = 1, and the test gives no information about convergence or divergence. Proof: (c) follows from considering the convergent series  1/n2 and the divergent series  1/n. In Example 4.1.11 we showed that lim n1/n = 1. Thus, Similarly, lim |1/n|1/n = 1, so the root test yields  = 1 for both series. Thus when  = 1 we can draw no conclusion about the convergence or divergence of a given series.  Note: This is very useful in applying the root test to other sequences.

(Integral Test) Theorem 8.2.13 Proof: Let f be a continuous function defined on [0, ), and suppose that f is positive and decreasing. That is, if x1 < x2, then f (x1)  f (x2) > 0. Then the series  f (n) converges iff exists as a real number. Theorem 8.2.13 Proof: Let an = f (n) and bn = Since f is decreasing, given any n  we have f (n + 1)  f (x)  f (n) for all x  [n, n + 1]. Since the length of each subinterval is 1, it follows that Geometrically, this means that the area under the curve y = f (x) from x = n to x = n + 1 is between f (n + 1) and f (n). Thus 0 < an +1  bn  an for each n. By the comparison test applied twice,  an converges iff  bn converges. y = f (x) But the partial sums of  bn are the integrals , so  bn converges precisely when exists as a real number.  Area is f (n + 1) Area is f (n) n n + 1

So, the p-series  1/n p converges if p > 1 and diverges if p  1. Any series of the form  1/n p , where p  , is called a p-series. Example 8.2.14 It is easy to see that the ratio and root tests both fail to determine convergence. But, we can use the integral test. For p  1, we have The limit of this as n   will be finite if p > 1 and infinite if p < 1. Thus by the integral test  1/n p converges if p > 1 and diverges if p < 1. When p = 1, we get the harmonic series, which is divergent. So, the p-series  1/n p converges if p > 1 and diverges if p  1. This can be useful when applying the comparison test.

(Alternating Series Test) If the terms in a series alternate between positive and negative values, the series is called an alternating series. Our final test gives us a simple criterion for determining the convergence of an alternating series. Theorem 8.2.16 (Alternating Series Test) If (an) is a decreasing sequence of positive numbers and lim an = 0, then the series  (1)n +1an converges. Proof: The idea of the proof is to show that the sequence (sn) of partial sums converges by considering two subsequences: the subsequence (s2n) coming from the sum of an even number of terms, and the subsequence (s2n+1) coming from the sum of an odd number of terms. In the text there are several examples of applying the various convergence tests to specific series.