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  1. $\displaystyle\sum_{n=0}^\infty \frac{z^{4n}}{n!}$
  2. $\displaystyle\sum_{n=0}^\infty n(n-1)z^n$

My thoughts:

  1. I think it'll look something like the exponential function but I'm not sure what exactly it would be.
  2. 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.
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You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Feb 23 '13 at 6:07
In q2 your index should start at n=1 – Daniel Littlewood Feb 23 '13 at 9:32

If the general term for the first question is meant to be $\dfrac{z^{4n}}{n!}$, then rewrite the term as $\dfrac{(z^4)^n}{n!}$, and note that we are looking at $e^{(z^4)}$.

For the second, consider the familiar $\sum_0^\infty z^n$, differentiate twice, and look at the result.

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And the series $\sum\limits_{n\geq0}n(n-1)z^n$ is the same as $z^2\sum\limits_{n\geq2}n(n-1)z^{n-2}$, which you should recognize.

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If you have done the differentiation theorem for power series, it can be used to evaluated number 2.

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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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