\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).
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.