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

Question

I want to compute the following infinite sum

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

To this goal, my strategy is to first compute $\sum_{n=2}^{\infty}\frac{1}{n^2-a^2}$, then take the derivative of the result with respect to $a$ and let $a \rightarrow 1$.

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

Recall the result

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

Hence

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

As a check, letting $a \rightarrow 1$ in $(1)$ I obtained

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

Which agrees with Wolfram.

Then, I differentiated $(1)$ w.r. to $a$

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

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

Letting $a=1$ in (2)

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

Edit, as pointed out by @user, $\psi^{\prime}(3)=\frac{\pi^2}{6}-\frac{5}{4}$ instead of $\psi^{\prime}(3)=\frac{\pi^2}{6}-2$, which gives now the right answer.

Answer 1

It should be

$$ \begin{aligned} \sum_{n=2}^{\infty}\frac{1}{(n^2-1)^2}&=\left(\frac{\psi^{\prime}(3)-\psi(3)}{4}+\frac{\psi^{\prime}(1)+\psi(1)}{4} \right)\\ &=\left(\frac{2\left(\frac{\pi^2}{6}-\color{red}{\frac54} \right)-2\left(-\gamma+\frac{3}{2} \right)}{8}+\frac{\frac{\pi^2}{3}-2\gamma}{8} \right)\\ &=\frac{\pi^2}{12}-\frac{11}{16} \end{aligned} $$

Answer 2

Here's a bit more Elementary and General proof ,

You need to compute $$\sum_{n>1}\frac{1}{(n^2-1)^2}\implies\sum_{n>1}\left(\frac{1}{(n-1)(n+1)}\right)^2\implies\sum_{n>0}\left(\frac{1}{n(n+2)}\right)^2$$

We've a sum like this i.e. $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^2=\frac{1}{k^2}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)$$

Here's why

Some basic concepts that I'll use $$\frac{1}{a\cdot b}=\frac{1}{b-a}\left(\frac{1}{a}-\frac{1}{b}\right)$$ and $$\color{red}{H_k^{(s)}=\sum_{n=1}^{k}\frac{1}{n^s}}$$

So , $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^2=\sum_{n=1}^{\infty}\frac{1}{k^2}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)^2$$ $$\Rightarrow\frac{1}{k^2}\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac{1}{(n+k)^2}-\frac{2}{n(n+k)}$$ $$=\frac{1}{k^2}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}+\color{}{\underbrace{\sum_{n=1}^{\infty}\frac{1}{(n+k)^2}+\color{red}{\sum_{n=1}^{k}\frac{1}{n^2}}}_{\zeta(2)}}-\color{red}{H_{k}^{(2)}}-\color{}{\sum_{n=1}^{\infty}\frac{2}{n(n+k)}}\right)$$ The second sum which we need to compute is $$\Rightarrow\sum_{n=1}^{\infty}\frac{1}{n(n+k)}=\sum_{n=1}^{\infty}\frac{1}{k}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)=\frac{1}{k}\left(\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{(n+k)}\right)$$ $$\Rightarrow \frac{1}{k}\sum_{n=1}^{k}\frac{1}{n}=\color{}{\frac{H_k^{(1)}}{k}}$$

Now finally assembling all the sums we get , $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^2=\frac{1}{k^2}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)$$

Now for $k=2$ you'll get $$\sum_{n>0}\left(\frac{1}{n(n+2)}\right)^2=\frac{1}{2^2}\left(2\zeta(2)-H_2^{(2)}-\frac{2H_2^{(1)}}{2}\right)=\frac{\pi^2}{12}-\frac{11}{16}$$

Answer 3

$$ \begin{aligned} \sum_{n=2}^{\infty} \frac{1}{\left(n^{2}-1\right)^{2}} &=\sum_{n=2}^{\infty}\left[\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right]^{2} \\ &=\frac{1}{4} \sum_{n=2}^{\infty}\left[\frac{1}{(n-1)^{2}}+\frac{1}{(n+1)^{2}}-\frac{2}{(n-1)(n+1)}\right] \\ &=\frac{1}{4}\left(\zeta(2)+\zeta(2)-1-\frac{1}{4}\right)-\frac{1}{4} \sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) \\ &=\frac{1}{4}\left(\frac{\pi^{2}}{3}-\frac{5}{4}\right)-\frac{3}{8} \\ &=\frac{\pi^{2}}{12}-\frac{11}{16} \end{aligned} $$