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.

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

Your example is very clear. I understand now why both conditions are necessary. thank you.