<s>[INST] 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.
 [/INST] 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 $c=2$
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