Find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$ find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$ $(a>0)$ 
I just can analyse $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}=a\left(\frac{1}{1}-\frac{1}{1+a}+\frac{1}{2}-\frac{1}{2+a}+\frac{1}{3}-\frac{1}{3+a}...+\frac{1}{n}-\frac{1}{n+a}\right)$ 
Can anyone help me? Thanks

It is $\psi (a + 1) + \gamma$, where $\psi$ is the logarithmic derivative of the gamma function and $\gamma$ is the Euler-Mascheroni constant, cf. http://dlmf.nist.gov/5.7.E6 and http://dlmf.nist.gov/5.5.E2 Using this fact, it follows for example that
$$
\log a + \gamma  + \frac{1}{{2a}} - \frac{1}{{12a^2 }} < \sum\limits_{n = 1}^\infty  {\frac{a}{{n(n + a)}}}  < \log a + \gamma  + \frac{1}{{2a}}
$$
for all $a>0$ (see http://dlmf.nist.gov/5.11.ii). Also, for $-1<a<1$, it holds that
$$
\sum\limits_{n = 1}^\infty  {\frac{a}{{n(n + a)}}} = \sum\limits_{k = 2}^\infty  {( - 1)^k \zeta (k)a^{k - 1} } ,
$$
where $\zeta$ denotes Riemann's zeta function (see http://dlmf.nist.gov/5.7.E4).
