Let $\sum \limits_{n=1}^{\infty} a_{n}$ be a convergent series when $\forall n, a_{n}\neq 0$.
Does $\sum \limits_{n=1}^{\infty} (1-\frac{\sin a_{n}}{a_{n}})$ converge if:
- $\forall n, a_{n}\gt 0$.
- $a_{n}\lt 0$ for infinitely many $n$'s.
For (1) I tried looking at the Maclaurian series for $\sin x$:
$|(1-\frac{\sin a_{n}}{a_{n}})|=|(1-\frac{1}{a_{n}}(a_{n}+R_{2}(a_{n}))|=|\frac{1}{6}a_{n}^2 \sin \xi| \leq a_{n}^2$
I'm not sure if what I did above is correct but if it is then by the comparison test we get that the series converges. Is there a simpler argument for this? Maybe something that uses the fact that $\sin a_{n} \leq a_{n}$?
Regarding (2) I think that it can converge or diverge. I tried to guess and it seems that $a_{n}=\frac{(-1)^n}{\sqrt n}$ yields a divergent series according to Mathematica but I'm not sure why. How can I find one that converges or diverges without guessing (maybe again by using a Maclaurian series)?