<s>[INST] Ramanujan's partial fraction decomposition of $\frac{1}{(x^2+a^2)\cdots(x^2+(a+n)^2)}$. \begin{align*}
\frac{1}{(x^2+a^2)\cdots(x^2+(a+n)^2)} &= \frac{2\Gamma(2a)}{\Gamma(n)\Gamma(2a+n)}\left(\frac{a}{x^2+a^2}-\frac{2a}{1!}\frac{n-1}{n+2a}\frac{a+1}{x^2+(a+1)^2}\right. \\
& \qquad + \left.\frac{2a(2a+1)}{2!}\frac{(n-1)(n-2)}{(n+2a)(n+2a+1)}\frac{a+2}{x^2+(a+2)^2}-\cdots\right).
\end{align*}
The preceding was by Ramanujan, appearing in one of his notebooks. How does one prove this? 
Especially interesting is motiving the proof:  given only the complete fraction on the left,  is there a method that makes the right side almost immediately obvious? (Basically, it would be nice if the answers imagined the RHS didn't exist in the above equation).
 [/INST] Since both the left- and right-hand sides are in terms of $x^2$ 
we can change variables to $X = x^2$.  This is a special case of
the question of finding the partial-fraction expansion
$$
\frac1{(X+A_1)(X+A_2)\cdots(X+A_n)}
= \sum_{i=1}^n \frac{C_i}{X+A_i}
$$
for any distinct $A_1,A_2,\ldots,A_n$.
The easy way to find $C_i$ is to multiply both sides by $X+A_i$
and then to evaluate at $X = -A_i$.  On the right side this isolates $C_i$.
On the left side we get the product over $j \neq i$ of $1/(A_j-A_i)$. 
So $C_i$ must equal this product.
In the present case, each $A_i$ is $(a+i)^2$,
so $A_j-A_i = (a+j)^2 - (a+i)^2$, which factors further as $(j-i)(2a+j+i)$.
The product of this over $1 \leq j \leq n$ excluding $j=i$ can then be
expressed in various ways in terms of factorials and Gamma functions,
one of which yields Ramanujan's choice.
</s>