# Sum of Power Series

• May 28th 2009, 12:56 PM
joeyjoejoe
Sum of Power Series
Compute $\displaystyle \sum_{0}^{\infty}(n^{2}+n)x^{n}$ for fixed values of x.

I know this sum is infinity when $\displaystyle \left|x \right|\geq1$, but how do you compute the value of the sum explicitly when $\displaystyle \left|x \right|<1$? The best I can seem to do is to use the Gaussian formula for the sum of the first n integers to deal with the $\displaystyle (n^{2}+n)$ term. Any suggestions?
• May 28th 2009, 01:38 PM
apcalculus
Quote:

Originally Posted by joeyjoejoe
Compute $\displaystyle \sum_{0}^{\infty}(n^{2}+n)x^{n}$ for fixed values of x.

I know this sum is infinity when $\displaystyle \left|x \right|\geq1$, but how do you compute the value of the sum explicitly when $\displaystyle \left|x \right|<1$? The best I can seem to do is to use the Gaussian formula for the sum of the first n integers to deal with the $\displaystyle (n^{2}+n)$ term. Any suggestions?

Integration by parts (integral test) test to find the two sums separately?

$\displaystyle \sum_{0}^{\infty}(n^{2}x^{n})$

and

$\displaystyle \sum_{0}^{\infty}(nx^{n})$

In the first sum you would have two use Parts twice. I hope this helps.
• May 28th 2009, 11:59 PM
Moo
Hello,

$\displaystyle (n^2+n)x^n=n(n+1) x^n$

Now rewrite this as $\displaystyle \frac 1x \cdot n(n+1) x^{n-1}$

And consider differentiating twice the power series $\displaystyle \frac{x}{1-x}=\sum_{n\geq 0} x^{n+1}$