Why is Part A Necessary in the Alternating Series Test?

[inkliing]

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.

 

[henry_m]

The assumption is necessary; a positive sequence need not be strictly decreasing to tend to zero. For example, the odd terms could be something like $$b_{2n-1}=1/n$$ and the even terms something like $$b_{2n}=1/n^2$$.

Think about the proof of the theorem. There are two ideas, one for each assumption:

The first idea, which requires your assumption (a), is that each term alternates in sign and is smaller than the previous term, and so the even partial sums form a decreasing sequence and the odd partial sums an increasing sequence, bounded below and above respectively. Each of these sequences of partial sums must tend to a limit. The sequence above fails here; in fact the sequence of odd partial sums will increase without bound.

The second idea is that if the limits of odd and even partial sums are equal, then the series converges to this common limit. This requires (b).

 

[inkliing]

That was very clear. I understand it now. Thank you.

 

[HallsofIvy]

It is NOT, strictly speaking, necessary that $b_{n+1}< b_n$ for all n, only that $b_n+1< b_n$ for "sufficiently large n". That is because you can always change the values of any finite number of terms of a series without changing whether it converges or not.