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I am trying to solve this problem:

For which $n>0$ the series $$\sum_{k=1}^{\infty}\frac{1}{k}\sin\left(\frac{k\pi}{n}\right)$$ converges?

My idea:

(0). Let $$a_k=\frac{1}{k}\sin\left(\frac{k\pi}{n}\right)$$ We have to check the convergence of $\sum a_k$.

(1). For $k\geq 0$ define $s_k$ as $$s_k=\sum_{j=1}^{n}a_{2kn+j}=\sum_{j=1}^{n}\frac{1}{2kn+j}\sin\left((2kn+j)\frac{\pi}{n}\right)$$ Then $s_k\geq 0$ as each summand $a_k\geq 0$. Also, we have $$s_k=\sum_{j=1}^{n}\frac{1}{2kn+j}\sin\left((2kn+j)\frac{\pi}{n}\right)\leq\sum_{j=1}^{n}\frac{1}{2kn+j}$$ Hence $s_k\rightarrow 0$ as $k\rightarrow \infty$.

(2). For $k\geq 0$ define $t_k$ as $$t_k=\sum_{j=n+1}^{2n}a_{2kn+j}=\sum_{j=n+1}^{2n}\frac{1}{2kn+j}\sin\left((2kn+j)\frac{\pi}{n}\right)$$ Then $t_k\leq 0$ as each summand $a_k\leq 0$. Also, we have $$|t_k|=\sum_{j=n+1}^{2n}\left|\frac{1}{2kn+j}\sin\left((2kn+j)\frac{\pi}{n}\right)\right|\leq\sum_{j=n+1}^{2n}\frac{1}{2kn+j}$$ Hence $|t_k|\rightarrow 0$ as $k\rightarrow \infty$.

(3). Notice that $|s_k|\geq |t_k|$ and $|t_k|\geq |s_{k+1}|$ for all $k$.

(4). Notice that $$\sum_{k=1}^{\infty}a_k=\sum_{k=0}^{\infty}(s_k+t_k)$$ From (1), (2) and (3) it follows that $\sum(s_k+t_k)$ satisfies the conditions for the alternating series test. Hence it is convergent.

Therefore the given series converges for all $n>0$.

I have two questions:

(1). Is my proof correct?

(2). I think that I am overthinking this one. Is there any other way to solve this problem?

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    $\begingroup$ (1), essentially. You're perhaps a bit short at point 3, just stating the result, but it's easy enough to see. However, as a card-carrying curmudgeon I have to note that the alternating series test only gives the convergence of a subsequence of the partial sums $\sum_{k = 1}^m a_k$, to conclude the convergence of the full sequence of the partial sums you need to say something more. (2), Dirichlet's test. It's made for this stuff. $\endgroup$ – Daniel Fischer Dec 27 '17 at 23:06
  • $\begingroup$ the proof is correct. Congrats! $\endgroup$ – Masacroso Dec 28 '17 at 0:20
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This looks correct.

A simpler way would be to prove that $|\sum_{k=1}^m\sin\frac {k\pi}{n}| $ is bounded ($m\in\mathbb N $). Actually, $\sum_{k=i}^{i+2n-1}\sin\frac {k\pi}{n}=0$ (any $2n$ consecutive terms add up to 0), so there is only up to $2n-1$ terms that don't get cancelled and the whole sum is then bounded by $2n-1$.

Now you can directly apply Dirichlet's convergence test to prove convergence.

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