Reducing Poisson Sum expression I am trying to reduce and obtain a simple expression from following sum
$$
\sum_{i = 1}^{\infty}{1 \over i}\,
{\,\mathrm{e}^{-\lambda}\,\lambda^{i} \over i!}\,
{1 \over 1 - \,\mathrm{e}^{-\lambda}\,}
$$
Problem can be put in general form as:
$$
\sum_{i = c}^{\infty}{1 \over i}\,
{\,\mathrm{e}^{-\lambda}\,\lambda^{i} \over
i!\left(\displaystyle{\, 1 - \sum_{k = 1}^{c - 1}
{\,\mathrm{e}^{-\lambda}\,\lambda^{k} \over k!}\,}\right)}\,,
\qquad  c \geq 1
$$

The first sum is
$$
e^{-\lambda}\int_0^\lambda\frac{e^x-1}{x}\,\mathrm{d}x
$$
which is not an elementary function. However, we can use the Exponential Integral to get
$$
e^{-\lambda}\sum_{k=1}^\infty\frac{\lambda^k}{k\,k!}=e^{-\lambda}\left(\mathrm{Ei}(\lambda)-\gamma-\log\left|\lambda\right|\right)
$$
