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I have this here:

Find the interval on which $\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n+3}$ is convergent and use the geometric series to find the sum of the series.

I already found the interval of convergence. It's $[-1,1)$. That wasn't too bad.

How do I use the geometric series to find the sum though? I know that $\displaystyle \sum_{n=0}^{\infty} x^n=1+x+x^2+x^3+...+x^n=\frac{1}{1-x}$ and I have to differentiate and integrate $\displaystyle \sum_{n=0}^{\infty}x^n$ as needed until it looks like $\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n+3}$ but how can I manipulate the $\displaystyle \sum_{n=0}^{\infty}x^n$ ? I am 99% sure I have to integrate in some way and shift indices but I am not sure.

Any help would be great!

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3 Answers 3

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We have $$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ and $$\frac{x^2}{1-x}=\sum_{n=0}^\infty x^{n+2}$$ by integrating $$\int\frac{x^2}{1-x}dx=\sum_{n=0}^\infty \int x^{n+2}dx\\ =\sum_{n=0}^\infty\frac{x^{n+3}}{n+3}\\ =x^3\sum_{n=0}^\infty\frac{x^{n}}{n+3}$$

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  • $\begingroup$ I follow what you did but the multiplication of the $x^2$ is something that I didn't expect to do haha. That's cool! $\endgroup$ Commented Feb 25, 2020 at 7:55
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If $f(x) = \sum\limits_{n=0}^{\infty} x^{n+2}$ then $\int_0^{x} f(t) dt= \sum\limits_{n=0}^{\infty} \frac {x^{n+3}} {n+3}=x^{3}\sum\limits_{n=0}^{\infty} \frac {x^{n}} {n+3}$. Can you continue from here?

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  • $\begingroup$ Yes I got it :) . It just seems unintuitive to just start from there. Thanks! $\endgroup$ Commented Feb 25, 2020 at 7:55
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$$S=\sum_{n=0}^{\infty}\frac{x^n}{n+3}\implies(x^3S)=\sum_{n=0}^{\infty}\frac{x^{n+3}}{n+3}\implies(x^3S)'=\sum_{n=0}^{\infty}{x^{n+3}}=x^3\sum_{n=0}^{\infty}{x^{n}}$$

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