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.
 A: $$\sum_{m\geq 0}\left(\frac{2}{3m+1}+\frac{1}{3m+2}-\frac{3}{3m+3}\right) = \sum_{m\geq 0}\int_{0}^{1}x^{3m}(2+x-3x^2)\,dx $$
equals:
$$ \int_{0}^{1}\frac{2+x-3x^2}{1-x^3}\,dx = \int_{0}^{1}\frac{2+3x}{1+x+x^2}\,dx = \color{red}{\frac{\pi}{6\sqrt{3}}+\frac{3\log 3}{2}}.$$
A: Hint:
This is the same as
$$\sum_{n=1}^{\infty}\left(\frac{2}{3n-2}+\frac{1}{3n-1}-\frac{3}{3n}\right)$$
A: Break the sum into three pieces
\begin{eqnarray*}
S=\sum_{n=0} ^{\infty} \left( \frac{2}{3n+1} + \frac{1}{3n+2} +\frac{-3}{3n+3} \right) \\
\int_0^1 x^n dx = \frac{1}{n+1} 
\end{eqnarray*}
Now use the above formula ... interchange the sum & integral & one has ... 
\begin{eqnarray*}
S= \int_0^{1} \frac{2+x-3x^2}{1-x^3} dx =\int_0^{1} \frac{2+3x}{1+x+x^2} dx =\frac{\pi}{6 \sqrt{3}} + \frac{3}{2} \ln 3. 
\end{eqnarray*}
A: 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 $0<x<1$.
