INFINITE SERIE 1/(n^2-1)^2

Today we will proof the following two infinte sums found here


\sum_{n=2}^{\infty}\frac{1}{(n^2-1)^2}=\frac{\pi^2}{12}-\frac{11}{16}


\sum_{n=2}^{\infty}\frac{n^2}{(n^2-1)^2}=\frac{\pi^2}{12}+\frac{1}{16}

First, lets evaluate the following infinite sum:


\begin{aligned} \sum_{n=2}^{\infty}\frac{1}{n^2-a^2}&=\frac{1}{2a}\sum_{n=2}^{\infty}\left(\frac{1}{n-a}-\frac{1}{n+a}\right)\\ &=\frac{1}{2a}\sum_{n=2}^{\infty}\left(\int_0^1t^{n-a-1}dt-\int_0^1t^{n+a-1}dt\right)\\ &=\frac{1}{2a}\sum_{n=2}^{\infty}\left(\int_0^1t^{-a}t^{n-1}dt-\int_0^1t^{a}t^{n-1}dt\right)\\ &=\frac{1}{2a}\left(\int_0^1t^{-a}\left(\sum_{n=2}^{\infty}t^{n-1}\right)dt-\int_0^1t^{a}\left(\sum_{n=2}^{\infty}t^{n-1}\right)dt\right)\\ &=\frac{1}{2a}\left(\int_0^1t^{-a}\left(\sum_{n=1}^{\infty}t^{n}\right)dt-\int_0^1t^{a}\left(\sum_{n=1}^{\infty}t^{n}\right)dt\right)\\ &=\frac{1}{2a}\left(\int_0^1t^{-a}\left(\frac{t}{1-t}\right)dt-\int_0^1t^{a}\left(\frac{t}{1-t}\right)dt\right)\\ &=\frac{1}{2a}\left(\int_0^1\frac{t^{1-a}}{1-t}dt-\int_0^1\frac{t^{a+1}}{1-t}dt\right)\\ &=\frac{1}{2a}\int_0^1\frac{t^{1-a}-t^{a+1}}{1-t}dt\\ \end{aligned}

Recall the following result

\int_0^1\frac{t^{z-1}-t^{w-1}}{1-t}dt=\psi(w)-\psi(z)

\sum_{n=2}^{\infty}\frac{1}{n^2-a^2}&=\frac{1}{2a}\left(\psi(a+2)-\psi(2-a) \right)(1)


For a=1 we obtain


\begin{aligned} \sum_{n=2}^{\infty}\frac{1}{n^2-1}&=\frac{1}{2}\left(\psi(3)-\psi(1) \right)\\ &=\frac{1}{2}\left(-\gamma+\frac{3}{2}+\gamma \right)\\ &=\frac{3}{4}\qquad \blacksquare\\ \end{aligned}

Lets now differentiate (1) w.r. to a


\begin{aligned} 2a\sum_{n=2}^{\infty}\frac{1}{(n^2-a^2)^2}&=\frac{\partial }{\partial a}\left(\frac{\psi(a+2)}{2a}-\frac{\psi(2-a)}{2a} \right) \\ &=\left(\frac{2a\psi^{\prime}(a+2)-2\psi(a+2)}{4a^2}+\frac{2a\psi^{\prime}(2-a)+2\psi(2-a)}{4a^2} \right) \\ \end{aligned}


\boxed{\sum_{n=2}^{\infty}\frac{1}{(n^2-a^2)^2}=\left(\frac{a\psi^{\prime}(a+2)-\psi(a+2)}{4a^3}+\frac{a\psi^{\prime}(2-a)+\psi(2-a)}{4a^3} \right)}(2)


Letting a=1 in (2)


\begin{aligned} \sum_{n=2}^{\infty}\frac{1}{(n^2-1)^2}&=\left(\frac{\psi^{\prime}(3)-\psi(3)}{4}+\frac{\psi^{\prime}(1)+\psi(1)}{4} \right)\\ &=\left(\frac{2\left(\frac{\pi^2}{6}-\frac{5}{4} \right)-2\left(-\gamma+\frac{3}{2} \right)}{8}+\frac{\frac{\pi^2}{3}-2\gamma}{8} \right)\\ &=\frac{\pi^2}{12}-\frac{11}{16} \qquad \blacksquare \end{aligned}

Now, for the second sum


\begin{aligned} \sum_{n=2}^{\infty}\frac{n^2}{(n^2-1)^2}&=\sum_{n=2}^{\infty}\frac{n^2-1+1}{(n^2-1)^2}\\ &=\sum_{n=2}^{\infty}\frac{n^2-1}{(n^2-1)^2}+\sum_{n=2}^{\infty}\frac{1}{(n^2-1)^2}\\ &=\sum_{n=2}^{\infty}\frac{1}{n^2-1}+\sum_{n=2}^{\infty}\frac{1}{(n^2-1)^2}\\ &=\frac{3}{4}+\frac{\pi^2}{12}-\frac{11}{16}\\ &=\frac{\pi^2}{12}+\frac{1}{16} \qquad \blacksquare \end{aligned}

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