inkliing Posted April 28, 2011 Share Posted April 28, 2011 There's something that's confusing me about what appears to be the standard form of stating the alternating series test in basic calculus. The four sources I looked up were James Stewart's CALCULUS, Howard Anton's CALCULUS, wolfram alpha's mathworld, and wikipedia. All four had essentially the same statement for the alternating series test: If the sum from i=0 to infinity of [(-1)^n][b_n], with b_n>0 for all n, satisfies a) b_(n+1)<=b_n for all n ({b_n} is a decreasing sequence), and b) the limit as n goes to infinity of b_n = 0, then the series is convergent. What I'm confused about is this: since all four sources made it clear that all of the b_n were strictly greater than zero and that b_n->0 as n goes to infinity, what is the point of also adding part a)? Why add that {b_n} must be monotonically decreasing (in the less-than-or-equal-to sense)? It seems to me that b_n>0 for all n and b_n->0 as n goes to infinity implies that {b_n} must be monotonically decreasing (in the less-than-or-equal-to sense). So it seems to me that all four sources should have left out the part about b_(n+1)<=b_n for all n. Am I missing something simple here? It seems to me that the needed counterexample is that of a sequence of real numbers, all greater than zero, which go to zero, but which are not eventually decreasing, which I'm pretty sure is impossible. I would appreciate it if some1 could clearly show that b), coupled with b_n>0 for all n, doesn't imply a) so I can be confident that it is necessary to state a) in the statement of the test. Thanks in advance. Link to comment Share on other sites More sharing options...

DrRocket Posted April 28, 2011 Share Posted April 28, 2011 It seems to me that the needed counterexample is that of a sequence of real numbers, all greater than zero, which go to zero, but which are not eventually decreasing, which I'm pretty sure is impossible. a_n = 1/n for n odd a_n=1/(2^n) for n even Link to comment Share on other sites More sharing options...

inkliing Posted April 30, 2011 Author Share Posted April 30, 2011 Your example is very clear. I understand now why both conditions are necessary. thank you. Link to comment Share on other sites More sharing options...

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