<s>[INST] Calculate $\sum_{n=1}^{\infty}{c_n\over n}$ when $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$ Let $(c_n)$ be defined as $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$. I want to get $\sum_{n=1}^{\infty}{c_n\over n}$. I would appreciate any suggestions how to start on this. It is my suspicion that I would have to resort to integration.
 [/INST] I'll expand on the (very clever!) method by Jack, giving some more details that may not have been obvious from his answer.
$$\begin{align}\sum_{n=1}^\infty\frac{c_n}{n}&=\sum_{k=0}^\infty\left(\frac{2}{3k+1} + \frac{1}{3k+2} - \frac{3}{3k+3}\right) \\
&=\sum_{k=0}^\infty\left(2 \int_0^1 x^{3k}dx + \int_0^1 x^{3k+1}dx - 3 \int_0^1 x^{3k+2}dx\right) \\
&=\sum_{k=0}^\infty \int_0^1 x^{3k}(2+x-3x^2) dx \\
&=\int_0^1 \sum_{k=0}^\infty x^{3k}(2+x-3x^2) dx \\
&=\int_0^1 \frac{2+x-3x^2}{1-x^3} dx \\
&=\int_0^1 \frac{2+3x}{1+x+x^2} dx \\
&=\int_0^1 \frac{\frac{3}{2}(2x+1) + \frac{1}{2}}{1+x+x^2} dx \\
&=\frac{3}{2}\int_0^1 \frac{2x+1}{1+x+x^2} dx +\frac{1}{2}\int_0^1 \frac{1}{1+x+x^2} dx\\
&=\frac{3}{2}\left[\log(1+x+x^2)\right]_0^1 +\frac{1}{2}\int_0^1 \frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx\\
&=\frac{3}{2} \log 3 +\frac{1}{2} \left[\frac{2}{\sqrt{3}}\arctan\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]_0^1 \\
&=\frac{3}{2} \log 3 +\frac{1}{\sqrt{3}}\left(\arctan \sqrt{3}-\arctan \frac{1}{\sqrt{3}}\right) \\
&=\frac{3}{2} \log 3 +\frac{\pi}{6\sqrt{3}} \\
\end{align}$$
Switching the integral and sum is justified by convergence of both involved quantities, and the geometric series is justified as $0<x<1$.
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