# Computation of a limit involving a series (related to Poisson distribution)

Consider $$\lambda >0.$$ I am reading a paper and the author states that

$$\displaystyle\lim_{v \rightarrow +\infty} \sum_{n=0}^{+\infty} \frac{\lambda^{n}}{(n !)^v} = 1 + \lambda$$

I tried to compute such limit but I am getting anywhere. Someone could help me?

Note that $$\sum_{n=0}^{+\infty} \frac{\lambda^{n}}{(n !)^v}=1+\lambda+\sum_{n=2}^{+\infty} \frac{\lambda^{n}}{(n !)^v}$$and $$\sum_{n=2}^{+\infty} \frac{\lambda^{n}}{(n !)^v}{\le \sum_{n=2}^{+\infty} \frac{\lambda^{n}}{n !}\sum_{n=2}^{+\infty} \frac{1}{(n !)^{v-1}}\\\le e^{\lambda}\sum_{n=2}^{\infty}{1\over 2^{(n-1)(v-1)}}\\=e^\lambda{{1\over 2^{v-1}}\over 1-{1\over 2^{v-1}}}\\={e^\lambda\over 2^{v-1}-1}\\\to 0}$$hence the result.
The RHS is obviously the first two terms of the sum. For the remaining terms, replace $$n!$$ by $$2^n$$. Then, whatever $$\lambda$$, for a sufficiently large $$v$$, you have a convergent geometric series (that tends to zero as $$v \to \infty$$).