Infinite summation I know that the sum $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$. How to find the sum of 
$$\sum_{n=1}^{\infty}\frac{1}{(a+bn)^{2}}$$
where $a,b>0$. (If $\frac{a}{b}$ is integer then the answer is direct. But what about general $a,b>0$?)
 A: In general, your infinite summation does not have a closed form. But if you introduce a special function which is reasonably famous and well-studied, you can find its value. Specifically, let
$$\psi_1 (s) = \frac{d^2}{ds^2} \log \Gamma(s)$$
be the trigamma function. Then we have
$$ \psi_1(s) = \sum_{n=0} \frac{1}{(s+n)^2}. $$
Therefore we can express your summation in terms of $\psi_1$:
$$ \sum_{n=1}^{\infty} \frac{1}{(a+bn)^2} = \frac{1}{b^2} \psi_1 \left( 1 + \frac{a}{b} \right).$$
In some special cases, you can find a closed form. For example, you already know that $\psi_1(1) = \zeta(2)$. Moreover, since
$$ \psi_1(z+1) = \psi_1(z) - \frac{1}{z^2},$$
we can concentrate on the case $0 \leq \frac{a}{b} < 1$. Then by
$$ \psi_1(1-z) + \psi_1(z) = \frac{\pi^2}{\sin^2 \pi z},$$
obtained by the log-differentiation of the Euler's reflection formula, we have
$$ \psi_1 \left(\frac{1}{2}\right) = \frac{\pi^2}{2},$$
which can also be obtained by some direct argument. Some other values for this function includes 
$$ \psi_1\left(\frac{1}{4}\right) = \pi^2 + 8K^2,$$
where $K$ is the Catalan constant.
A: As an alternative to the trigamma function proposed by sos440 you may use the Hurwitz zeta function :
$$\zeta(s,q)=\sum_{n=0}^\infty \frac 1{(q+n)^s}$$
so that $$\frac 1{b^2}\zeta\left(2,1+\frac ab\right)= \sum_{n=1} \frac{1}{(a+b\,n)^2}. $$
