Hello i need to determine the radius of convergense for the following series:

$\sum_{n=0}^{\infty}\frac{1}{\sqrt{n!}}x^n$

Im thinking the ratio test, but as im pretty uncertain about these types of problems, i'd like a second opinion, and a bit of help

2. Yes, for any "radius of convergence" problem, the best way to go is either the ratio test or the root test (and most of the time the ratio test is much simpler than the root test).

Here, $a_n= \frac{1}{\sqrt{n!}}x^n$ so $a_{(n+1)}= \frac{1}{\sqrt{n+1}}x^{n+1}$.

Then $\left|\frac{a_{n}}{a_{n+1}}\right|= \frac{\sqrt{n!}}{\sqrt{(n+1)!}}|x|$ $= \sqrt{\frac{n!}{(n+ 1)!}}|x|$

What is the limit of that as n goes to infinity?
For what x is that limit less than 1?

3. The limit is 1, so x < 1 will make the limit less than 1 making the radius of convergence = 1 ?

4. Originally Posted by Zaph
The limit is 1, so x < 1 will make the limit less than 1 making the radius of convergence = 1 ?
note that $\frac{n!}{(n+1)!} = \frac{1}{n+1}$ ... care to rethink your response?

5. Originally Posted by skeeter
note that $\frac{n!}{(n+1)!} = \frac{1}{n+1}$ ... care to rethink your response?
hmm, $\frac{1}{n+1}$ diverges right?.. Im lost

6. Originally Posted by Zaph
hmm, $\frac{1}{n+1}$ diverges right?.. Im lost
$\frac{1}{n + 1} \to \frac{1}{\infty} \to 0$ as $n \to \infty$.

So what do you think the radius of convergence has to be?

7. Aha! So the radius of convergence is infinity or R? since the series goes to 0.

8. Originally Posted by Zaph
Aha! So the radius of convergence is infinity or R? since the series goes to 0.
Yes, since this limit is $0$, it is always $<1$.

So the series converges for all $x$.

9. Thanx a bunch everyone, im really struggling with these types of problems, but its getting better and better