Edit: Thanks for the help everyone. Turns out I was just making an arithmetic error. Whoops!
Question answered
I've got this infinite series which starts at k=2
$$\sum_{k=2}^{\infty} \frac 1 {k^2-1}$$
https://www.wolframalpha.com/input/?i=infinite+sum+1%2F(k%5E2-1)
I thought this series might be telescoping, so I decomposed it to: 1/2(k-1) - 1/2(k+1). Then I wrote out partial sums and found that after k=4, terms begin to cancel.
Then I summed up the parts that don't cancel and got the sum = 7/12.
However according to wolfram, the sum is 3/4. I don't understand what I'm missing.
If following my work in text form is confusing, I will gladly write it out and upload a picture.
Let $$f(x)=\sum_{k=2}^{\infty} \frac 1 {k^2-1}x^{k-1}$$ then one has \begin{eqnarray} f'(x)=\sum_{k=2}^{\infty} \frac 1 {k+1}x^{k-2},(x^3f'(x))'=\sum_{k=2}^{\infty}x^{k}=\frac{x^2}{1-x}. \end{eqnarray} Thus \begin{eqnarray} f(1)&=&\int_0^1\frac{1}{t^3}\int_0^t\frac{x^2}{1-x}dxdt\\ &=&\int_0^1\int_0^t\frac{x^2}{t^3(1-x)}dxdt\\ &=&\int_0^1\int_x^1\frac{x^2}{t^3(1-x)}dtdx\\ &=&\int_0^1\frac{x^2}{1-x}\int_x^1 t^{-3}dtdx\\ &=&\frac12\int_0^1(1+x)dx\\ &=&\frac34. \end{eqnarray}