# Expressing series in terms of a Power Series

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1. Jul 18, 2017

### pkmpad

Hello and thank you for trying to help.

In spite of the fact that this seems a very simple problem, I do not find myself able to get a solution. Here it goes:

Let $$f(x)=\displaystyle \sum_{k=3}^\infty a_k \frac{x^k}{k(k-1)(k-2)}$$ and $$g(x)=\displaystyle \sum_{k=0}^\infty a_k x^k$$. Express f(x) in terms of g(x).

My attempt has been to express f(x) as
$$\displaystyle\sum_{k=3}^{\infty}a_k\frac{x^k}{k(k-1)(k-2)}=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-\sum_{k=3}^{\infty}\frac{a_k x^k}{k-1}+\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k-2}$$

And then

$$f(x)=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-x\sum_{k=2}^{\infty}\frac{a_k x^k}{k}+\frac{x^2}{2}\sum_{k=1}^{\infty}\frac{a_k x^k}{k}$$

What should I do next? I suppose that I have to integrate those series, but I have already tried that and I have not been able to get any relationship with g(x).

Any help will be appreciated.

Last edited: Jul 18, 2017
2. Jul 18, 2017

### scottdave

Your LaTex is not displaying correctly. I have tried 2 different browsers. It looks like you have f(x) = Sum of (k=3 to infinity) (ak xk) / (k(k-1)(k-2)),
then g(x) = sum of (k=0 to infinity) (ak xk), then I get confused trying to follow your workings, after that.

3. Jul 18, 2017

### pkmpad

Corrected, thank you

4. Jul 18, 2017

### Ray Vickson

Using summation index $j = k-2$ your sum becomes
$$f(x) = \sum_{j=1}^{\infty} a_{j+2} \frac{x^{j+2}}{j(j+1)(j+2)}$$

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