Closed form for the sum: $\sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)}$ I tried using partial fractions to compute the sum of the series
$$
\sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)}
$$
Another technique is to turn this series into a definite integral of 0 to 1. but do not know how to do.  
Thanks for any help.
 A: Note that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n(n+1/3)} = \sum_{n=1}^{\infty}\dfrac{9}{3n(3n+1)} = 9\sum_{n=1}^{\infty}\left(\dfrac{1}{3n}-\dfrac{1}{3n+1}\right)$. 
Now, let $f(x) = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{x^{3n}}{3n}-\dfrac{x^{3n+1}}{3n+1}\right)$. Clearly, $f(0) = 0$ and our sum is $9f(1)$. 
Also, termwise differentiation yields $f'(x) = \displaystyle\sum_{n=1}^{\infty}\left(x^{3n-1}-x^{3n}\right) = \dfrac{x^2-x^3}{1-x^3} = \dfrac{x^2}{1+x+x^2}$. 
Using partial fractions, we get $f(1) = f(0) + \displaystyle\int_{0}^{1}f'(x)\,dx$ $= \displaystyle\int_{0}^{1}\dfrac{x^2}{1+x+x^2}\,dx = \left[x - \dfrac{1}{2}\ln(1+x+x^2) - \dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right) \right]_0^1$ $= 1 - \dfrac{1}{2}\ln 3 - \dfrac{\pi}{6\sqrt{3}}$
Therefore, our sum is $9f(1) = 9 - \dfrac{9}{2}\ln 3 - \dfrac{\pi\sqrt{3}}{2}$.
A: You can start with this identity

$$ \sum_{n=1}^{\infty}\frac{x^{n-2/3}}{n} = -\frac{\ln(1-x)}{x^{2/3}}. $$

Now just integrate both sides w.r.t. $x$ from $0$ to $1$ to get the answer

$$ -\pi \,\sqrt {3}/2+9-9\,\ln  \left( 3 \right)/2.  $$

Note: We used the identity 

$$ \sum_{n=1}^{\infty}\frac{x^{n}}{n}=-\ln(1-x). $$

which is easy to prove.
A: Hint
$\frac{1}{(n)(n+\frac{1}{3})}=3(\frac{1}{n}-\frac{1}{n+\frac{1}{3}})$
