How to evaluate this infinite series? $\sum_{r = 1}^{\infty} \frac{1}{r(r + \frac{1}{3})}$ The question is to find this infinite series:
$$ \sum_{r = 1}^{\infty} \frac{1}{r(r + \frac{1}{3})} = \; ? $$

I have tried applying the Riemann sum to the Integral technique as follows:
Assuming the lower and upper boundaries (a and b) are 0 and 1. Now, we know that
$$ \int_{0}^{1} {f(x)} \; dx = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} f(\frac{r}{n}) $$
Therefore,
$$ \sum_{r = 1}^{\infty} \frac{1}{r(r + \frac{1}{3})} = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{r(r + \frac{1}{3})} $$
$$ = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{n}{r(r + \frac{1}{3})} = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{n}{r} \frac{1}{({r+\frac{1}{3}})} $$
I'm stuck here.
Is there a different approach to this problem? Or, could this problem be solved this way?

Edit: Given in this comment below, does not answer my question. The answer is completely incomprehensible to me.
 A: Using partial fractions, we have $$\dfrac{1}{r(r+\tfrac{1}{3})} = \dfrac{3}{r}-\dfrac{3}{r+\tfrac{1}{3}}.$$ Then, by using the identity $\dfrac{1}{n} =  \displaystyle\int_{0}^{1}x^{n-1}\,dx$, we have $$S:= \sum_{r = 1}^{\infty}\dfrac{1}{r(r+\tfrac{1}{3})} = \sum_{r = 1}^{\infty}\left[\dfrac{3}{r}-\dfrac{3}{r+\tfrac{1}{3}}\right] = 3\sum_{r = 1}^{\infty}\int_{0}^{1}\left(x^{r-1}-x^{r-2/3}\right)\,dx.$$ Because every term is positive, we can then interchange the order of the summation and integral. $$S = 3\int_{0}^{1}\sum_{r = 1}^{\infty}\left(x^{r-1}-x^{r-2/3}\right)\,dx.$$ The summation is of a geometric series which converges for $|x| < 1$ (i.e. almost everywhere on $[0,1]$), so we have $$S = 3\int_{0}^{1}\dfrac{1-x^{1/3}}{1-x}\,dx.$$ Now, just substitute $x = u^3$, $dx = 3u^2\,du$ and use a trigonometric substitution to finish.
A: The logarithmic derivative of the Gamma function, a famous special function generalizing the factorial operations to complex inputs, is called the digamma function, $\psi$. For complex $z$, it can be shown (from the Weierstrass factorization of the gamma function) that
$$\psi(z+1) = - \gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$$
So note
$$\sum_{n=1}^{\infty} \frac{z}{n(n+z)} = \psi(z+1)+\gamma$$
and so
$$\sum_{n=1}^{\infty} \frac{\frac{1}{3}}{n(n+\frac{1}{3})} = \psi(\frac{4}{3})+\gamma$$
Thus
$$\sum_{n=1}^{\infty} \frac{1}{n(n+\frac{1}{3})} = 3(\psi(\frac{4}{3})+\gamma)$$
A: Making the problem more general, for $a>0$, we can compute the partial sum
$$S_n=\sum_{r=1}^n \frac 1{r(r+a)}$$ Using partial fraction decomposition
$$S_n=\frac 1 a  (-\psi (a+n+1)+\psi (a+1)+\psi (n+1)+\gamma )$$ and using the asymptotics
$$S_n=\frac{\psi(a+1)+\gamma }a-\frac{1}{n}+\frac{ (a+1)}{2 n^2}-\frac{ (a+1) (2 a+1)}{6
   n^3}+O\left(\frac{1}{n^4}\right)$$
