Closed expression for infinite series How can I find the value of the series $$\sum_{n=1}^\infty \frac{1}{(b+n)(a+n)}$$
where $a,b\in[0,1)$?
It is obvious by standard arguments that the series converges, but how can I derive the explicit value dependent on $a$ and $b$.
Thanks
EDIT: $a,b$ can be assumed to be different.
 A: The whole key is generalization! :-) Let us first notice that for natural values of a and b, we 
have $~\dfrac1{(a+n)(b+n)}=\dfrac1{b-a}\bigg(\dfrac1{a+n}-\dfrac1{b+n}\bigg),~$ turning our sum into a telescoping series, 
whose value becomes $~\dfrac{H_a-H_b}{a-b}~,~$ where $~H_n=\displaystyle\sum_{k=1}^n\frac1k~$ for $n\in$ N. But what if $n\not\in$ N ? :-$)$ 
Let us notice that $~\displaystyle\sum_{k=0}^{n-1}x^k=\frac{1-x^n}{1-x~~}~,~$ which, by integrating both sides between $x=0$ and 
$x=1$, yields $~H_n=\displaystyle\int_0^1\frac{1-x^n}{1-x~~}dx,~$ which expression, unlike the abovementioned formula 
for harmonic numbers, can easily be evaluated for non-natural arguments as well, and, more 
than that, when a and b are fractional rationals, possesses a closed form; e.g., for $n=\dfrac pq~,~$ 
we simply substitute $t=\sqrt[q]x,~$ and then employ partial fraction decomposition and/or other 
available tools which can be used for determining the primitives of rational functions. This 
approach has been discovered by Euler some two and a half centuries ago. A list of exact 
values for $H_n$ evaluated at fractional arguments can be found here. Hope this helps.
