# Thread: Series and Convergence

1. ## Series and Convergence

Hey,

The question ask you to find all values of x for which the series converges. For these values of x, write the sum of the series as a function of x.

$\displaystyle\large\sum\limits^\infty_{n=0}(-1)^{n}x^{2n}$

It is alternating, so if the ratio is r = -x^2 then -x^2 must be less than 1 to converge, and if it r = x^2 then x^2 must be less than 1 to converge.

This is where I got lost. It seems i'll have 2 different answers. How would I be able to come up with the sum of the series as a function afterward?

Help?

2. By the ratio test, x must be between -1 and 1. The endpoints x=-1 and x=1 must be checked separately. Both give divergent series by the divergence test. So the series converges if and only if $-1.

The series is geometric with first term 1, and common ratio $-x^2$. So the sum is $\frac{1}{1+x^2}$

3. If you recognise that this is a geometric series with $\displaystyle a = 1$ and $\displaystyle r = -x^2$, you know that infinite geometric series only converge for $\displaystyle |r| < 1$.

So $\displaystyle |-x^2| < 1$

$\displaystyle |x|^2 < 1$

$\displaystyle |x| < 1$

$\displaystyle -1 < x < 1$.

Then you don't need to resort to using the ratio test and checking endpoints.