I have a double sum function as \begin{align*} F = \sum^{\infty}_{i=1} \sum^{\infty}_{j=1} \frac{(i \times j)\lambda^{i+j}}{(i+j)!} \end{align*}
I tried to transform it to the form of $\sum^{\infty}_{n=2}f(n)\frac{\lambda^{n}}{n!}$ to apply the Power Series Expansion for Exponential Function, but could not find the pattern for $f(n)$.
Could you give me some hints about how to conduct the transform? Or the approximation for the function $F$? I'm very appreciated it.
You can calculate this explicitly collecting equal powers of $\lambda$.
We have
$$\sum_{i+j=n;i,j\geq 1}ij = \sum_{i=1}^{n-1}i(n-i)=n\sum_{i=1}^{n-1}i -\sum_{i=1}^{n-1}i^2$$
$$= n\cdot\frac{n(n-1)}{2}-\frac{(n-1)n(2n-1)}{6}$$ $$= \frac{n^3-n}{6}$$
Hence, $$ \sum^{\infty}_{i=1} \sum^{\infty}_{j=1} \frac{(i \times j)\lambda^{i+j}}{(i+j)!} = \frac 16\sum_{n=2}^{\infty}(n^3-n)\frac{\lambda^n}{n!} $$