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