I have a function as \begin{align*} F = \sum^{\infty}_{i=1} \sum^{\infty}_{j=1} \frac{i!j!}{(i+j)!} \left(e^{-\lambda} \frac{\lambda^{i}}{i!}\right)\left(e^{-\lambda} \frac{\lambda^{j}}{j!}\right) \end{align*}
I want to find the exact value of F or a way to approximate it.
In fact, by removing the duplicate components, F is similar to \begin{align*} F = e^{-2\lambda}\sum^{\infty}_{i=1} \sum^{\infty}_{j=1} \frac{\lambda^{i+j}}{(i+j)!} \end{align*}
This form is pretty similar to the Power Series Expansion for Exponential Function. However, the double sum makes me confused.
Could anyone give me a hint about it? Thank you very much.
You can calculate this explicitly by collecting equal powers of $\lambda$.
To do so, note that $i+j = n$ for $n\geq 2$ has exactly $n-1$ possible positive integer solutions.
Hence, calculating only the double sum without the factor $e^{-2\lambda}$:
$$\sum^{\infty}_{i=1} \sum^{\infty}_{j=1} \frac{\lambda^{i+j}}{(i+j)!} = \sum^{\infty}_{n=2} (n-1)\frac{\lambda^{n}}{n!}$$ $$= \lambda \sum^{\infty}_{n=2} \frac{\lambda^{n-1}}{(n-1)!} - \sum^{\infty}_{n=2}\frac{\lambda^{n}}{n!}$$ $$= \lambda(e^{\lambda}-1) - (e^{\lambda}-1-\lambda)$$
I leave it up to you to multiply by $e^{-2\lambda}$.