How is the given summation containing exponential function $\sum_{a=0}^{\infty} \frac{a+2} {2(a+1)} X \frac{(a+1){(\lambda X)}^a e^{-\lambda X}}{a! (1+\lambda X)}=\frac{X}{2} (1+ \frac{1}{1+\lambda X})$
I tried the following.
$=\sum_{a=0}^{\infty} \frac{a+2} {2(a+1)} X \frac{(a+1){(\lambda X)}^a e^{-\lambda X}}{a! (1+\lambda X)}$
$=\sum_{a=0}^{\infty} \frac{a+2} {2} X \frac{{(\lambda X)}^a e^{-\lambda X}}{a! (1+\lambda X)}$
$=\sum_{a=0}^{\infty} \frac{X(a+2)} {2(1+\lambda X)} \frac{{(\lambda X)}^a e^{-\lambda X}}{a! }$
$=\sum_{a=0}^{\infty} \frac{X(a+2)} {2(1+\lambda X)} \sum_{a=0}^{\infty} \frac{{(\lambda X)}^a e^{-\lambda X}}{a! }$
As $\sum_{a=0}^{\infty} \frac{{(\lambda X)}^a e^{-\lambda X}}{a! }=1$, the sum of poisson pmf sum up to 1.
$=\frac{X} {2(1+\lambda X)} \sum_{a=0}^{\infty}(a+2) $
I don't know how to further proceed. Can someone explain me the remaining steps to get to the answer. thanks in advance.
After removal of clutter, the summation is essentially
$$\sum_{k=0}^\infty\frac{k+2}{k!}t^k=t\sum_{k=0}^\infty\frac{1}{(k-1)!}t^{k-1}+2\sum_{k=0}^\infty\frac{1}{k!}t^k=(t+2)e^t$$ where $t=\lambda X$. The remaining factors are $\dfrac{Xe^{-t}}{2(t+1)}$.