Can we argue that as $\sum^\infty _{n=m} a_n$ is $\sum^\infty _{n=m}|(-1)^n a_n|$, $\sum^\infty _{n=m}(-1)^n a_n$ is convergent?

Question

I want to prove

Let $(a_n)^\infty_{n=m} $ be a sequence of real numbers which are non negative and decreasing, thus $a_n\geq 0$ and $a_n\geq a_{n+1}$ for evere $n \geq m.$ Then the series $\sum^\infty _{n=m}(-1)^n a_n$ is convergent iff the sequence $a_n$ converges to $0$ as $n\rightarrow \infty$.

From Zero test it follows that $a_n$ converges to $0$. Hence $\sum^\infty _{n=m} a_n$ converges.
Can we argue that as $\sum^\infty _{n=m} a_n$ is $\sum^\infty _{n=m}|(-1)^n a_n|$, $\sum^\infty _{n=m}(-1)^n a_n$ is convergent due to absolute convergence test?
I am asking this question since the book I am following -Analysis 1 by Tao gives a longer proof for this statement.

Answer 1

Yes, you can (answer to the yellow part)...but it is rather trivial, since:

(i) If $\;a_n\to0\;$ then we have all the conditions of Leibniz series and thus the series converges, and

(ii) If the series converges then the series sequence must converge to zero, so $\;a_n\to0\;$

By "the series" I mean above , as you wrote:

$$\sum_{n=m}^\infty (-1)^na_n$$

What you wrote after the yellow part isn't correct. See comments.

Answer 2

Classical counter example: $a_n=\frac1n$. Converges to zero, but the absolute series does not converge. However, the alternating series converges to $\ln 2$.

It is true that in general absolute convergence, i.e., $\sum_{n=1}^\infty |a_n| <\infty$ implies convergence of the series with signs, independent of how the signs are distributed. Indeed $\sum_{n=1}^\infty a_n·b_n$ converges for any bounded sequence $(b_n)$.