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).

First of all there is a error in the expression on the left hand side. The correct right hand side is:
\begin{equation}
lhs=\frac{1}{(x^2+a^2)\dot \cdots \dot (x^2+(a+n-1)^2)}
\end{equation}
The left hand side is correct.
Now from partial fraction decomposition the right hand side clearly equals:
\begin{equation}
rhs = \sum\limits_{k=0}^{n-1} \frac{1}{x^2+(a+k)^2} \cdot C_k 
\end{equation}
where:
\begin{eqnarray}
C_k&=&\frac{1}{\prod\limits_{j=0,j\neq k}^{n-1} (-(a+k)^2 + (a+j)^2)}=\frac{1}{\prod\limits_{j=0,j\neq k}^{n-1} (2 a+k+j)(j-k)} \\
&=&\frac{(-1)^{k}}{(2a+k)^{(k)} (k)! (2a+2 k+1)^{(n-1-k)}(n-1-k)!}\\
&=& \frac{1}{\Gamma(n)} \cdot \binom{n-1}{k} \cdot (-1)^k \cdot \frac{\Gamma(2a+k)}{\Gamma(2 a+n+k)} (2a+2 k) \\
&=& \frac{2 \Gamma(2a)}{\Gamma(n) \Gamma(2 a+n)} \cdot \binom{n-1}{k} (-1)^k \cdot \frac{(2 a)^{(k)}}{(2a+n)^{(k)}} \cdot (a+k)
\end{eqnarray}
which is exactly what we have on the right hand side. Here of course $k=0,\cdots,n-1$.
