In my last exam there was a task where we had to find the value of this series:
$$ \sum_{k=0}^\infty \frac{n^{2k}}{k!} $$
I thought about using a ratio test where I did the following:
$$ \dfrac{a_{k+1} }{a_{k}} = \frac{\dfrac{n^{2*(k+1)}}{(k+1)!}}{\dfrac{n^2*k}{k!}} = \dfrac{n^{2(k+1)}k!}{(k+1)!*n^{2k}} $$
The problem that I have now is that I don't know how to continue or if this the right approach.
Hint:
$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$
Your expression is $$\sum_{k=0}^\infty \frac{n^{2k}}{k!}=\sum_{k=0}^{\infty}\frac{(n^2)^k}{k!}$$