<s>[INST] I have to find the marginal pmf's $f_X$, $f_Y$ and $f_{X+Y}$ of $X+Y$. I have to find $D > 0$ such that $f(x, y)$ = $D$($\frac{1}{x+y−1}$$+$ $\frac{1}{x+y+1}$ $−$ $\frac{2}{x+y}$) is the joint pmf $f_{X,Y} (x, y)$ of a random vector $(X,Y)$ in 
{$1,2,..$}$^2$. And then I have to find the marginal pmf's $f_X$, $f_Y$ and $f_{X+Y}$ of $X+Y$.
My attempt: First I tried to find $D$
$\sum_{x=1}^{\infty}$$\sum_{y=1}^{\infty}$ $D$($\frac{1}{x+y−1}$$+$ $\frac{1}{x+y+1}$ $−$ $\frac{2}{x+y}$) = $1$ 
D$\sum_{x=1}^{\infty}$ $\sum_{y=1}^{\infty}$($\frac{2}{(x+y−1)(x+y+1)(x+y)})$ = $1$
$D$$\sum_{x=1}^{\infty}$ $\frac{1}{x^2+x}$= $1$
$D$ = $1$
Then:
$f_X(x)$ = $\sum_{y=1}^{\infty}$$D$($\frac{2}{(x+y−1)(x+y+1)(x+y)})$
$f_X(x)$ = $\sum_{y=1}^{\infty}$$\frac{2}{(x+y−1)(x+y+1)(x+y)}$
$f_X(x)$ = $\frac{1}{x^2+x}$
Then:
$f_Y(y)$ = $\sum_{x=1}^{\infty}$$D$($\frac{2}{(x+y−1)(x+y+1)(x+y)})$
$f_Y(y)$ = $\sum_{x=1}^{\infty}$$\frac{2}{(x+y−1)(x+y+1)(x+y)}$
$f_Y(y)$ = $\frac{1}{y^2+y}$
For $f_{X+Y}$ I was thinking of the following:
Set $U=X$ and $V=X+Y$, then use the change of variables theorem, note that:
$\begin{pmatrix}
\frac{\partial U}{\partial X} & \frac{\partial U}{\partial Y} \\
\frac{\partial V}{\partial X} & \frac{\partial V}{\partial Y} 
\end{pmatrix}$
= $\begin{pmatrix}
1 & 0 \\
1 & 1 
\end{pmatrix}$ has determinant one. Hence,
$f_{U,V}$ = $f_{X,Y}(x(u,v),y(u,v))$ = $\frac{2}{(x+y−1)(x+y+1)(x+y)}$ = $\frac{2}{(u+v-u-1)(u+v-u+1)(u+v-u)}$ = $\frac{2}{(v-1)(v+1)v}$ by the change of variables
Now we get via $f_V$
$f_V(v)$ = $\sum_{v=2}^{\infty}$ $\frac{2}{(v-1)(v+1)v}$ = $\frac{1}{2}$
Now I have to calculate the conditional pmf's $f_{X|Y}(x|y)$ and $f_{Y|X}(y|x)$
I know that we have to use:
$f_{X|Y}(x|y)$ = $P(X=x|Y=y)$ = $\frac{P(X=x,Y=y)}{P(Y=Y)}$
I am not sure if the above answers are correct. 
So any help would be grateful. Thanks in advance.
 [/INST] You are correct concerning $D$, $f_X$ and $f_Y$.
For finding $f_{X+Y}$ realize that for $n=2,3,\dots$ we have:
$P\left(X+Y=n\right)=\sum_{x+y=n}\left[\frac{1}{x+y-1}+\frac{1}{x+y+1}-\frac{2}{x+y}\right]=\left(n-1\right)\left[\frac{1}{n-1}+\frac{1}{n+1}-\frac{2}{n}\right]$
</s>