TITLE: Find the corresponding functions of the following power series
QUESTION [2 upvotes]: $\displaystyle\sum_{n=0}^\infty \frac{z^{4n}}{n!}$
$\displaystyle\sum_{n=0}^\infty n(n-1)z^n$

My thoughts:

I think it'll look something like the exponential function but I'm not sure what exactly it would be.
the hint says to divide by $z^2$, and I got the power series to look like $$2z^2+6z^3+12z^4+20z^5+\cdots,$$ but I'm not sure where to go from there.

REPLY [0 votes]: 1) we know that $$e^x=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}$$ let $x=z^4$ and you will get your first series.2) you have the following$$\displaystyle \sum_{n=0}^{\infty}n(n-1)z^n=0+0+\displaystyle \sum_{n=2}^{\infty}n(n-1)z^{n-2}=\displaystyle \sum_{n=2}^{\infty}n(n-1)z^{n-2}$$ but we also have$$\frac{1}{1-z}=\displaystyle \sum_{n=0}^{\infty}z^n$$ if you differentiate twice you'll get the result. Remark: for part (2) the result is true if $|z|<1$.
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