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Assume that $a_n≥0$ for all $n$ and that $\sum_{n=1}^{\infty}a_n<\infty$. Must $\sum_{n=1}^{\infty}(-1)^na_n$ also converge?

Intuitively, I believe this should be true. By the n-term test, it must be true that the sequence a_n must converge to zero for the series to converge (which is also a requirement in the Alternating series test). However, is it necessarily true that $a_{n+1}≤ a_n$ for all $n$?

If not, may you provide a counterexample?

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  • $\begingroup$ Are you assuming that $a_n≥0$? $\endgroup$ – lulu Apr 3 at 19:41
  • $\begingroup$ Yes, a_n is supposed to always be non-negative $\endgroup$ – Allan Lago Apr 3 at 19:42
  • $\begingroup$ That should be added to the post. I'll do that for you, and reformat at the same time, though it is a good idea to learn how to format for the site...here's a good tutorial. $\endgroup$ – lulu Apr 3 at 19:43
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    $\begingroup$ Yes, in fact it converges absolutely. One of the basic theorems about convergence of infinite series says that if $\Sigma|c_n|$ converges then $\Sigma c_n$ converges. If $a_n\ge0$ and $c_n=(-1)^na_n$ then $|c_n|=a_n$. Hint: use Cauchy's criterion. $\endgroup$ – bof Apr 3 at 19:46
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Yes, and this can be easily shown the following way:

$$\sum_{n=1}^\infty (-1)^n a_n = \sum_{n=1}^\infty (1+(-1)^n - 1)a_n = \sum_{n=1}^\infty (1+(-1)^n)a_n - \sum_{n=1}^\infty a_n$$

The above is valid because the series

$$\sum_{n=1}^\infty (1+(-1)^n)a_n \le 2\sum_{n=1}^\infty a_n$$

converges.


As for the second part of your question: It is not necessary that $a_n \ge a_{n+1}$ for all $n$. This isn't even true if you replace "for all $n$" with "for sufficiently large $n$". As a counter-example, consider the positive sequence $$a_n = \frac{1}{n} + \frac{(-1)^n}{2n} = \begin{cases} \dfrac{1}{2n}, & \text{for } n \text{ odd} \\ \dfrac{3}{2n}, & \text{for } n \text{ even} \end{cases}$$

For $n$ odd, we have

$$a_n = \frac{1}{2n},\ a_{n+1} = \frac{3}{2(n+1)},\ a_{n+1}>a_n$$

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