# Use the geometric series to find the sum of the series

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!

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 $$\int\frac{x^2}{1-x}dx=\sum_{n=0}^\infty \int x^{n+2}dx\\ =\sum_{n=0}^\infty\frac{x^{n+3}}{n+3}\\ =x^3\sum_{n=0}^\infty\frac{x^{n}}{n+3}$$
• 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! – Future Math person Feb 25 '20 at 7:55
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?
$$S=\sum_{n=0}^{\infty}\frac{x^n}{n+3}\implies(x^3S)=\sum_{n=0}^{\infty}\frac{x^{n+3}}{n+3}\implies(x^3S)'=\sum_{n=0}^{\infty}{x^{n+3}}=x^3\sum_{n=0}^{\infty}{x^{n}}$$