Try This Sum

The adjacent solution is great and perfect for the problem so great like this.

Anyways here is how I did it

$\displaystyle S = \sum_{n=2}^{\infty}{\frac{{e}^{-2n}}{{n}^{2}-1}}$

What the function under the sigma operation can remind? First of all, it can create a lot of tension for sure.

Can it be written in the form of a Laurent series and then supposedly Residue Theorem? Well, I don't think so. Then, some other integration under the sum sign?

Yeah. Laplace transform!

$\displaystyle {L}^{-1}\left(\frac{{e}^{-2n}}{{n}^{2}-1}\right) = {u}_{2}(x)sinh(x-2)$

as it is of the form $\displaystyle {e}^{-cs} F(s)$ , $\displaystyle F(s) = \frac{m}{{s}^{2}-{m}^{2}}$ and $\displaystyle s = n, m = 1$

where $\displaystyle {L}^{-1}$ means inverse Laplace transform, $\displaystyle {u}_{c}(x)$ means heaviside step function. For finding the inverse Laplace transform, one can either peek it from the table or the Bromwich integral(although latter is quite hard to do).

Hence,

$\displaystyle {L}^{-1}(S) = \sum_{n=2}^{\infty}{{u}_{2}(x)sinh(x-2)}$

$\displaystyle S = \sum_{n=2}^{\infty}{\int_{0}^{\infty}{{u}_{2}(x)sinh(x-2){e}^{-nx}} dx}$

Using GP formula,

$\displaystyle S = \int_{0}^{\infty}{{u}_{2}(x) \frac{{e}^{-x}}{{e}^{x}-1} sinh(x-2) dx}$

which is just

$\displaystyle S = L\left(\frac{{u}_{2}(x)sinh(x-2)}{{e}^{x}-1}\right)$

equals

$\displaystyle S = \frac{1}{2}\left(1 + \frac{{e}^{-2}}{2}\right) + sinh(2)ln(1-{e}^{-2})$

Before we start evaluating this sum let's simplify it into an easier one: $S\quad =\quad \sum _{ n=2 }^{ \infty }{ \frac { { e }^{ -2n } }{ { n }^{ 2 }-1 } } \\ \quad \quad =\quad \frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty }{ (\frac { { e }^{ -2n } }{ n-1 } } \quad -\quad \frac { { e }^{ -2n } }{ n+1 } )\\ \quad \Rightarrow 2S\quad =\quad S'-S''$

Let's simplify $S'$ and $S''$ :

$S'\quad =\quad \sum _{ n=2 }^{ \infty }{ \frac { { e }^{ -2n } }{ n-1 } } \quad =\quad { e }^{ -2 }\sum _{ n=2 }^{ \infty }{ \frac { { e }^{ -2(n-1) } }{ n-1 } } \\ \qquad \qquad \qquad \qquad \quad =\quad { e }^{ -2 }\sum _{ k=1 }^{ \infty }{ \frac { { e }^{ -2k } }{ k } } \quad =\quad { e }^{ -2 }I$ And : $S''\quad =\quad \sum _{ n=2 }^{ \infty }{ \frac { { e }^{ -2n } }{ n+1 } } \quad =\quad { e }^{ 2 }\sum _{ n=2 }^{ \infty }{ \frac { { e }^{ -2(n+1) } }{ n+1 } } \\ \qquad \qquad \qquad \qquad \quad =\quad { e }^{ 2 }\sum _{ k=3 }^{ \infty }{ \frac { { e }^{ -2k } }{ k } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad { e }^{ 2 }(\sum _{ k=1 }^{ \infty }{ \frac { { e }^{ -2k } }{ k } } -\quad ({ e }^{ -2 }+\frac { { e }^{ -4 } }{ 2 } ))\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad { e }^{ 2 }(I-({ e }^{ -2 }+\frac { { e }^{ -4 } }{ 2 } ))$

Now let's evaluate $I$ : We consider the following generating function $y\quad =\quad f(x)\\ \quad \quad =\quad \sum _{ k=1 }^{ \infty }{ \frac { { e }^{ -2k } }{ k } } { x }^{ n }\\ \quad \quad =\quad x{ e }^{ -2 }+\sum _{ k=1 }^{ \infty }{ \frac { { e }^{ -2(k+1) } }{ k+1 } } { x }^{ k+1 }\\ \Rightarrow y'\quad =\quad { e }^{ -2 }+\sum _{ k=1 }^{ \infty }{ { e }^{ -2(k+1) } } { x }^{ k }\\ \quad \quad y'\quad =\quad { e }^{ -2 }+{ e }^{ -2 }\sum _{ k=1 }^{ \infty }{ { e }^{ -2k } } { x }^{ k }$ And : $y\quad =\quad f(x)\\ \quad \quad =\quad \sum _{ k=1 }^{ \infty }{ \frac { { e }^{ -2k } }{ k } } { x }^{ n }\\ \Rightarrow y'\quad =\quad \sum _{ k=1 }^{ \infty }{ { e }^{ -2k } } { x }^{ k-1 }\\ \quad xy'\quad =\quad \sum _{ k=1 }^{ \infty }{ { e }^{ -2k } } { x }^{ k }$

So, we get the following differential equation and it's solution :

$y'={ e }^{ -2 }+{ e }^{ -2 }xy'\\ y'(1-x{ e }^{ -2 })={ e }^{ -2 }\\ y'=\frac { { e }^{ -2 } }{ 1-x{ e }^{ -2 } } \\ \Rightarrow y\quad =\quad -ln(1-x{ e }^{ -2 })+C$

We can verifiy that for $x=0$ , $y=0$ , So $C=0$ :

$I=f(1)=-ln(1-{ e }^{ -2 })$

Now we can find the first Sum:

$2S={ I(e }^{ -2 }-{ e }^{ 2 })+(1+\frac { { e }^{ -2 } }{ 2 } )$ And by substituting $I$ , we get:

$S=ln(1-{ e }^{ -2 })\frac { { e }^{ 2 }-{ e }^{ -2 } }{ 2 } +\frac { 1 }{ 2 } (1+\frac { { e }^{ -2 } }{ 2 } )\\ S=ln(1-{ e }^{ -2 })sinh(2)+\frac { 1 }{ 2 } (1+\frac { { e }^{ -2 } }{ 2 } )$