Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  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.
share|improve this question
    
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
add comment

4 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

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 $|z|<1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.