Convergence of $\sum_{ k\in\mathbb{N} } \left( \frac{\lambda^k}{k!} \right)^n$

Question

We know that $\sum_{ k\in\mathbb{N} } \frac{\lambda^k}{k!} = e^\lambda$. I'm interested in the convergence of $$S^{(n)}=\sum_{ k\in\mathbb{N} } \left( \frac{\lambda^k}{k!} \right)^n $$ for some value of $n$. The series general term $a_k= \left( \frac{\lambda^k}{k!} \right)^n$ converges for $k\geq 1$. Applying the comparison test of this series with exponential series $$\lim_{k\rightarrow+\infty} \frac{\left( \frac{\lambda^k}{k!} \right)^n}{\frac{\lambda^k}{k!}} = 0$$ , we may conclude that the sequence converges; i.e. the series converges. Numerically i tested the behavior for $n\rightarrow+\infty$ of $S^{(n)}$ and the series converges to $2$.
Now, if there is no mistake in what I wrote before, I wan't to know if there is any way to calculate the value of convergence $S$? Is there any relation between $S$ and $e$ or $e^\lambda$?

Answer 1

Let $$ S^{(n)}(\lambda)=\sum_{k=0}^\infty\frac{\lambda^{nk}}{(k!)^n}. $$ Then $$ \lim_{n\to\infty}S^{(n)}(\lambda)=\begin{cases} 0 & \text{if }0\le\lambda<1,\\ 2 & \text{if }0\le\lambda=1,\\ \infty &\text{if }\lambda>1. \end{cases} $$ Let's prove it. If $0\le\lambda<1$ then $$ 1\le S^{(n)}(\lambda)=1+\sum_{k=1}^\infty\frac{\lambda^{nk}}{(k!)^n}\le1+\sum_{k=1}^\infty\lambda^{nk}=1+\frac{\lambda^n}{1-\lambda^n}. $$ If $\lambda>1$ then $$ \le S^{(n)}(\lambda)\ge 1+\lambda^n. $$ Finally, since $k!\le2^k$ if $k\ge4$, we have $$ 2\le S^{(n)}(1)\le2+\frac{1}{2^n}+\frac{1}{6^n}+\sum_{k=4}^\infty\frac{1}{2^{nk}}=2+\frac{1}{2^n}+\frac{1}{6^n}+\frac{2^{-4n}}{1-2^{-n}}. $$