I would like to calculate the following series, and I don't really know where to begin, or how to do it.
$\sum _{n=0}^{\infty }\:\frac{\left(nj^n+1\right)}{n!}$
$\sum _{n=0}^{\infty }\:\frac{n^2j^n}{\left(2n\right)!}$
$\sum _{n=0}^{\infty }\:\frac{1}{\left(kn\right)!}$ with k being from N*
I am not asking for complete solutions (It would be appreciated). I want a starting point or some ideas on how to do these kind of exercises.
j is the imaginary unit ($j^2=−1$)
For the first series.Firstly you can easily prove that the series is convergent (by comparison). So you can decompose the series in a sum of two convergent series and add their respective limits.
$$\sum _{n=0}^{\infty }\:\frac{\left(nj^n+1\right)}{n!}=\sum _{n=0}^{\infty }\:\frac{nj^n}{n!}+\sum _{n=0}^{\infty }\:\frac{1}{n!}=\sum _{n=1}^{\infty }\:\frac{nj^n}{n!}+e=j\sum _{n=0}^{\infty }\:\frac{(n+1)j^n}{(n+1)!}+e$$
The first series is the derivative of : $$\sum _{n=0}^{\infty }\:\frac{j^n}{n!}=e^j$$ So $$j\sum _{n=0}^{\infty }\:\frac{(n+1)j^n}{(n+1)!}=j(e^j)'=je^j$$
Finally : $$\sum _{n=0}^{\infty }\:\frac{\left(nj^n+1\right)}{n!}=je^j+e$$
For the second series you can use the power series of $\sinh$ and $\cosh$.
For the third series you will probably need to use some kind of hypergeometric function