# If a positive terms series converges, must the alternating series of the same form also converge?

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?

• Are you assuming that $a_n≥0$? – lulu Apr 3 at 19:41
• Yes, a_n is supposed to always be non-negative – Allan Lago Apr 3 at 19:42
• 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. – lulu Apr 3 at 19:43
• 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. – bof Apr 3 at 19:46

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

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