So we have this series: $$ \sum^{\infty}_{n = 1}\left(-1\right)^n\sin\left(a \over n\right) $$
The task is to find all $a$ values, in which case the series converges and all $a$ values in which case the series absolutely converges.
My first question is that is this even an alternating series ?. From my maths book: It says that series is an alternating series $\sum^{\infty}_{n = 1}\left(-1\right)^{n - 1}a_n$, where $a_{n} > 0\,,\ \forall\ n\in N$. But $\sin\left(a/n\right)$ is from $-1$ to $1$.
So I know that the series $\sum^{\infty}_{n = 1}u_{n}$ converges absolutely if $\sum^{\infty}_{n = 1}\left\vert u_{n}\right\vert < \infty$.
So I tried using D'Alemberts formula $\lim_{n \to \infty}\left\vert u_{n+1}/u_{n}\right\vert = D$ to find the series convergence, but got $D = 1$, so this is no good.
But I don't think these take into account the $a$ variable, how can I find out what the actual $a$ values need to be ?. Should I just start trying different values for $a$ ?. Any suggestions, tips would be greatly appreciated !.
Whatever value $a$ has, over the long term $a/n$ will be very close to $0$, and so $\sin \frac an\sim\frac an$. So the series might not be alternating for the first terms, but it will always be eventually.
The series only converges absolutely for $a=0$. For $a\ne0$, the series does converge, but not absolutely.
To see this, we can assume $a>0$ (the case $a<0$ is identical, since $\sin x=-\sin(-x)$. For $n$ big enough, we have $\sin \frac an\geq\frac a{2n}$, so $$ \sum_n\left|\sin\frac an\right|\geq\sum_n\frac a{2n}=\infty. $$ We also have $\sin \frac an\to0$ and, for $n$ big enough $\sin \frac an>0$. So by the Leibniz Criterion, the series converges.