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Prove/disprove: We have a positive sequence $a_n$ such that $\displaystyle\lim_{n\to\infty}a_n=0$ so the series $\displaystyle\sum^\infty_{n=1} (-1)^na_n $ converge.

It's very similar to Leibniz alternating sum test but I think the statement is false so I'm trying to find a counter example. If there's a positive sequence that tends to zero but isn't monotone decreasing then the alternating sum test won't work, thus the series does not converge.

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The alternating series test only gives a suficient condition for convergence. – Zircht Apr 29 '14 at 16:31
Take, e.g., $a_n=1/n^2$, $n$ odd; $a_n=1/n$, $n$ even. – David Mitra Apr 29 '14 at 16:33
@ᛥᛥᛥ you mean there can be an alternating series that doesn't satisfy the alternating sum test but still converge ? – GinKin Apr 29 '14 at 16:34
@GinKin Yes. Try to find a convergent series similar to the one given by David Mitra. – Zircht Apr 29 '14 at 16:40
@ᛥᛥᛥ $a_n=1/n^2$, $n$ is odd, $a_n=0$, $n$ is even. I see. – GinKin Apr 29 '14 at 16:42
up vote 2 down vote accepted

This community wiki solution is intended to clear the question from the unanswered queue.

First off, failing the alternating series test doesn't imply that the series doesn't converge.

The example you gave is perfect: $\displaystyle a_n = \begin{cases}1/n^2 &\text{if $n$ is odd}, \\ 0 & \text{if $n$ is even}. \end{cases}$.

Notice that $\displaystyle \lim_{n \to \infty} a_n = 0$ but $a_n$ is not monotonically decreasing (it jumps up and down from $1/n^2$ to $0$).

However $\displaystyle \sum_{k = 1}^\infty a_n < \sum_{k = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} < \infty$ and hence converges.

Now, to disprove the statement in question, define $\displaystyle a_n = \begin{cases}1/n^2 &\text{if $n$ is odd}, \\ 1/n & \text{if $n$ is even}. \end{cases}$.

It is immediate that $\displaystyle \lim_{n \to \infty} a_n = 0$.

Notice that $$\displaystyle \sum_{n \text{ odd}} a_n = \sum_{n \text{ odd}} \frac{1}{n^2}< \sum_{k = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ and $$\displaystyle \sum_{n \text{ even}} a_n = \sum_{n \text{ even}} \frac1n = \frac12 \sum_{k = 1}^\infty \frac1n = \frac12 \cdot \infty = \infty.$$ Therefore $$\displaystyle \sum_{n = 1}^\infty (-1 )^n a_n = \sum_{n \text{ even}} a_n - \sum_{n \text{ odd}} a_n = \infty - \text{ finite positive number} = \infty.$$

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