Here are some relatively short questions that I wanted quick answers to:
If there exists $\sum_{n=0}^{\infty}a_n$ that converges absolutely, then $\sum_{n=0}^{\infty}(-1)^nn a_n$ converges.
My answer: yes, for example, let $a_n=\frac{1}{n^2}$, then $(-1)^n \frac{1}{n}$ does converge, albeit conditionally.
Second, if there is a $a_n >0 \forall n \in \mathbb{N}$ which $\sum_{n=1}^{\infty}a_n$ converges, then $\sum_{n=1}^{\infty}\sin(a_n)$ converges.
My answer: yes also. By the divergence test, $\lim_{n \to \infty}a_n=0$ for sure. Then $\lim_{n \to \infty}\sin(0)=0$. But this does not exactly prove that $a_n$ converges, only that its limit is 0...Can someone check if this is correct?
Consider the sequence
$a_n = \begin{cases} \frac{1}{n^2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$
Then $\sum\limits_{n = 0}^\infty a_n$ converges absolutely, but $\sum\limits_{n = 0}^\infty (-1)^n n a_n$ diverges.
For the second case, we see that if $a_n \geq 0$ for all $n$ and $\sum\limits_{n = 0}^\infty a_n$ converges, $\sum\limits_{n = 0}^\infty \sin(a_n)$ converges as well.
This is because for sufficiently large $n$, we must have $0 < a_n < \frac{\pi}{2}$ since we know that $\lim\limits_{n \to \infty} a_n = 0$. Thus, for sufficiently large $n$, we have $\sin(a_n) > 0$ and $\sin(a_n) < a_n$. Thus, $\sum\limits_{n = 0}^\infty \sin(a_n)$ converges by the comparison test.