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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 $a,b>0$?)

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2 Answers 2

up vote 4 down vote accepted

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.

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+1: For other fractions you may rewrite $\psi_1(f)$ in function of the Clausen functions (see here). –  Raymond Manzoni Oct 27 '12 at 14:43
    
@sos440: Thank you so much! –  Jessica Oct 27 '12 at 17:07
    
@RaymondManzoni, thanks for pointing out that. Now I remember that the Clausen functions and polylogarithms are helpful when evaluating zeta-like series such as $\psi_2(\frac{1}{3})$. –  sos440 Oct 27 '12 at 17:52

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 $$\frac 1{b^2}\zeta\left(2,1+\frac ab\right)= \sum_{n=1} \frac{1}{(a+b\,n)^2}. $$

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