Double summation equation involving a parameter While trying to solve a statistics probability density function question, I have reached a point where I unable to proceed with solving for the value of $c$ in the equation:
$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac{c}{(i+j-1)(i+j)(i+j+1)} = 1$, where $c>0$
I went through the list of series and sequences, including Taylor series, but I could not find any which will help simplify the summations and solve for the value of $c$. How can I solve such an equation? I would really appreciate some help with this.
 A: HINT
Consider the term in the double summation, $$\frac{c}{(i+j-1)(i+j)(i+j+1)} = \frac{c((i+j+1)-(i+j-1))}{2(i+j-1)(i+j)(i+j+1)}$$ $$ = \frac{c}{2(i+j)}.(\frac{1}{(i+j-1)} - \frac{1}{i+j+1})$$ $$= \frac{c}{2}(\frac{1}{i+j-1} - \frac{2}{i+j} + \frac{1}{i+j+1})$$
Now , you fix $k = i+j$$$\implies S = \sum_{k=0}^\infty \frac{c}{2}(\frac{1}{k-1} - \frac{2}{k} + \frac{1}{k+1})$$
But, we need to consider the number of times each sum would appear, i.e, how many times $k$ takes a certain value, corresponding to different pairs of $(i,j)$ . We observe that for any number $n$, we can express it as sum of   two whole numbers less than it in exactly $n+1$ ways (Try this out for yourself, proof is trivial). So, we modify the sum above to 
$$S = \sum_{k=0}^\infty \frac{c}{2}(\frac{k+1}{k-1} - \frac{2(k+1)}{k} + 1) = 1$$
A: The starting point is partial fraction decomposition :
$$\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2(i+j-1)}-\frac{1}{i+j}+\frac{1}{2(i+j+1)}$$
Now, let's compute a partial sum (indexed by $j$) :
$$\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\sum_{j=1}^q\left(\frac{1}{i+j-1}-\frac{1}{i+j}\right)-\frac{1}{2}\sum_{j=1}^q\left(\frac{1}{i+j}-\frac{1}{i+j+1}\right)$$
After re-indexation :
$$\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\sum_{k=i+1}^{i+q}\left(\frac{1}{k-1}-\frac{1}{k}\right)-\frac{1}{2}\sum_{k=i+1}^{i+q}\left(\frac{1}{k}-\frac{1}{k+1}\right)$$
that is :
$$\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\left(\frac{1}{i}-\frac{1}{i+q}\right)-\frac{1}{2}\left(\frac{1}{i+1}-\frac{1}{i+q+1}\right)=\frac{1}{2}\left(\frac{1}{i}-\frac{1}{i+1}\right)-\frac{1}{2}\left(\frac{1}{i+q}-\frac{1}{i+1+q}\right)$$
Now the second summation :
$$\sum_{i=1}^p\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\sum_{i=1}^p\left(\frac{1}{i}-\frac{1}{i+1}\right)-\frac{1}{2}\sum_{i=1}^p\left(\frac{1}{i+q}-\frac{1}{i+1+q}\right)$$
that is :
$$\sum_{i=1}^p\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\left(1-\frac{1}{p+1}\right)-\frac{1}{2}\left(\frac{1}{q+1}-\frac{1}{p+q+1}\right)$$
Taking the limit as $p,q\to\infty$, we get :
$$\sum_{i=1}^\infty\sum_{j=1}^\infty\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}$$
Therefore, you should choose $c=2$
A: Put $r=i+j$.  
For  $i,j=1 \rightarrow \infty$, we have  $r=2\rightarrow \infty$, and the number of times $r$ occurs is $\color{blue}{(r-1)}$. 
Hence
$$\begin{align}
1&=\sum_{i=1}^\infty\sum_{j=1}^\infty\frac c{(i+j-1)(i+j)(i+j+1)}&&\text{(given)}\\
&=\sum_{r=2}^\infty \color{blue}{(r-1)}\cdot \frac c{(r-1)r(r+1)}\\
&=c\sum_{r=2}^\infty \frac 1{r(r+1)}\\
&=c\sum_{r=2}^\infty \frac 1r-\frac 1{r+1}\\
&=c\cdot \frac 12 &&\text {(sum by telescoping)}\\
\Rightarrow c&=\color{red}2
\end{align}$$
