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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.

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    Have you tried partial fraction decomposition? – 2017-01-19
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    Looks like the sum oh that series depends on $c$ and $j$ ... Are you sure that it isn't a double sum like $\sum_{i=1}^\infty\left(\sum_{j=1}^\infty\cdots\right)$ ? – 2017-01-19
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    @Adren, especially considering the title says "Double summation...." – 2017-01-19
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    My apologies, it is a double sum. I completely forgot to add the second one. – 2017-01-19
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    @Adren I have added this now. I would appreciate some help with this – 2017-01-19

3 Answers 3

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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

$$\begin{align} 1&=\sum_{i=1}^\infty\sum_{j=1}^\infty\frac c{(i+j-1)(i+j)(i+j+1)}&&\text{(given)}\\ &=\sum_{r=2}^\infty \color{blue}{(r-1)}\cdot \frac c{(r-1)r(r+1)}\\ &=c\sum_{r=2}^\infty \frac 1{r(r+1)}\\ &=c\sum_{r=2}^\infty \frac 1r-\frac 1{r+1}\\ &=c\cdot \frac 12 &&\text {(sum by telescoping)}\\ \Rightarrow c&=\color{red}2 \end{align}$$

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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 $$S = \sum_{k=0}^\infty \frac{c}{2}(\frac{k+1}{k-1} - \frac{2(k+1)}{k} + 1) = 1$$

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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 $c=2$