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
 A: 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).
A: Let's call the original sum $\lim_{n \to \infty} V_n$.
Asymptotic solution for $V_n$ with $a>0$: the first sum is Harmonic, so it is $\log n + O(1)$. The second sum is 
$$
\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k+a} = \lim_{n \to \infty} S_n
$$
Each value in the (argument, value) tuple in this sum, $(1, \frac{1}{1+a}), (2, \frac{1}{2+a}) \ldots (n, \frac{1}{n+a})$ is in fact an area of a rectangle: $r_1 = (2-1) \times \frac{1}{1+a}, r_2= (3-2) \times \frac{1}{2+a} , \ldots r_n =  (n+1-n) \times \frac{1}{n+a}$, so the sum $S_n$ is equa; to the sum of areas of these rectangles. 
Next step is to compare each $r_j$ to the function $f(x) = \frac{1}{x+a}$. For each interval $[1,2], (2,3], \ldots (n, n+1)$ area of $r_j$ upper-bounds integral of $f(x)$:
$$
r_j > \int_{j}^{j+1} f(x)dx = \log \frac{j+1+a}{j+a}
$$
If we sum LHS and RHS of this inequality, we get the lower-bound on S_n:
$$
S_n > \sum_{j=1}^{n} > \sum_{j=1}^{n} \log \frac{j+1+a}{j+a} = \log (n+a+1) - \log (a+1)
$$ 
As a result, you get an upper bound on the original sum:
$$
V_n < H_n - \log (n+a+1) + \log (a+1) = \log (a+1) + \gamma + \log (\frac{n}{n+a+1}) = \log (a+1) + \gamma + O(\frac{1}{n})
$$
EDIT: got the sign wrong the first time. Also $\log \frac{n}{n+a+1} = -\log (1+\frac{a+1}{n}) \sim - \frac{a+1}{n} = O(\frac{1}{n})$ 
