# Thread: Finding values for which a sequence converges..

1. ## Finding values for which a sequence converges..

For which values of x does $\displaystyle \sum_{n=0}^\infty \frac{x^{2n}}{\sqrt{n+1}}$ converge? Justify your answer.

Now, I reckon the sequence converges for -1>x<1, because anything else looks like it will just diverge... but how do I go about "justifying" this?

2. Originally Posted by Unenlightened
For which values of x does $\displaystyle \sum_{n=0}^\infty \frac{x^{2n}}{\sqrt{n+1}}$ converge? Justify your answer.

Now, I reckon the sequence converges for -1>x<1, because anything else looks like it will just diverge... but how do I go about "justifying" this?
Let $\displaystyle S_n = \frac{x^{2n}}{\sqrt{n+1}}$

You have to find the radius (and hence the interval) of convergence for this power series.

You can use the ratio test. Start by finding $\displaystyle \mathop {\lim }\limits_{n \to \infty } | \frac{S_{n+1}}{S_n}|= \frac{|x|^{2n+1}}{\sqrt{n+2}} / \frac{|x|^{2n}}{\sqrt{n+1}}$

3. root test also works here.