Sum of the form $\sum_{ij}^\infty a_ia_j 2^{\min(i,j)}$ I want to find 
$$
S:= \sum_{i,j = 0}^\infty a_ia_j 2^{\min(i,j)}
$$
where $a_j=\lambda^j/j!$ for some fixed $\lambda>0$.  
I’m pretty sure this converges because $2^{\min(i,j)} \leq 2^{i +j}$ so
$$
S \leq \sum_{ij}^\infty a_ia_j 2^{i +j} = \left( \sum_j (2\lambda)^j/j!\right)^2 = e^{4\lambda}
$$
but I don’t know what to do with this actual sum. Thanks for any help. 
 A: Let's assume $N\le M$, then $$\sum_{i=0}^N \sum_{j=0}^M a_ia_j 2^{\min(i,j)}=\sum_{i=0}^N \left(\sum_{j=0}^i a_i a_j 2^j + \sum_{j=i+1}^M a_i a_j 2^i\right)=\\
\sum_{i=0}^N \frac{\lambda^i}{i!} \left( \sum_{j=0}^i \frac{\lambda^j}{j!} 2^j + 2^i \sum_{j=i+1}^M \frac{\lambda^j}{j!}\right) $$
Now $$\sum_{j=0}^i \frac{\lambda^j}{j!} 2^j=e_i(2\lambda)$$ 
where $e_i(x)$ is the incomplete exponential sum function
So we get
$$
\sum_{i=0}^N \frac{\lambda^i}{i!} \left(e_i(2\lambda)+2^i (e_M(\lambda)-e_i(\lambda))\right)=\\e_M(\lambda)e_N(2 \lambda)+ \sum_{i=0}^N \frac{\lambda^i}{i!} \left( e_i(2\lambda)- 2^i e_i(\lambda)\right)$$
I'm not sure if we can evaluate this for $N,M\to \infty$. The first term tends to $e^{3\lambda}$, but the sum needs some more work.
The sum, equivalently, can be seen as an event/sum over a joint distribution of two independent Poisson r.v (with parameters $\lambda$ and $2 \lambda$). 
Specifically:
$$ S=\sum_{i=0}^\infty \frac{\lambda^i}{i!} \frac{(2 \lambda)^i}{i!}+ 2 \sum_{i=0}^\infty \sum_{j=0}^{i-1} \frac{\lambda^i}{i!} \frac{(2 \lambda)^j}{j!}  $$
Not, let $X\sim Poisson(\lambda)$ , $Y\sim Poisson(2 \lambda)$  (independent). Then the summand above are:
$$ S_1=e^{3\lambda} P(Y=X)$$
$$ S_2=e^{3\lambda} P(Y<X)$$
Unfortunately it's not easy to compute those probabilities, see for example here and here.
A: $$
S:= \sum_{i,j = 0}^\infty a_ia_j 2^{\min(i,j)}=\sum_{i=0}^\infty \frac{\lambda^i}{i!}\left(\sum_{j=0}^i \frac{\lambda^j}{j!}2^j+\sum_{j=i+1}^\infty \frac{\lambda^j}{j!}2^i\right)
$$
$$
=\sum_{i=0}^\infty \frac{\lambda^i}{i!}\left(\frac{e^{2\lambda}}{i!}\int_{2\lambda}^\infty dt\ e^{-t}t^{i}+2^i\sum_{j=0}^\infty\frac{\lambda^j}{j!}-2^i\sum_{j=0}^i\frac{\lambda^j}{j!}\right)
$$
$$
=e^{2\lambda}\int_{2\lambda}^\infty dt e^{-t}\sum_{i=0}^\infty\frac{(\lambda t)^i}{(i!)^2}+\sum_{i=0}^\infty\frac{(2\lambda)^i}{i!}\sum_{j=0}^\infty\frac{\lambda^j}{j!}-\sum_{i=0}^\infty\frac{(2\lambda)^i}{i!}\frac{e^\lambda}{i!}\int_\lambda^\infty dt e^{-t}t^i
$$
$$
=\boxed{e^{2\lambda}\int_{2 \lambda }^{\infty } e^{-t} I_0\left(2 \sqrt{t\lambda} \right) \, dt+e^{3\lambda}-e^{\lambda}\int_\lambda^\infty dt\ e^{-t}I_0\left(2 \sqrt{2\lambda t} \right)}\ ,
$$
where $I_0$ is a Bessel function. The integrals are seemingly not expressible in closed form. 
