# Thread: stuck in a step for finding radius and interval of convergence

1. ## stuck in a step for finding radius and interval of convergence

find the interval and radius of convergence for sum from n=1 to infinity of $\frac{x^n}{\sqrt{n}}$. So I used the ratio test and I've got it down to $|x| \lim_{n\to\infty}|\sqrt{\frac{n}{n+1}}|$. I want to use L'Hospitals rule but I'm uncertain how to proceed. Can I use L'H rule on $\frac{n}{n+1}$ or do I use it on $\frac{\sqrt{\frac{n}{n+1}}}{1}$

2. no need for L'Hopital

$\sqrt{\frac{n}{n+1}} = \sqrt{\frac{1}{1 + \frac{1}{n}}}
$

now take the limit as $n \to \infty$

3. Originally Posted by superdude
find the interval and radius of convergence for sum from n=1 to infinity of $\frac{x^n}{\sqrt{n}}$. So I used the ratio test and I've got it down to $|x| \lim_{n\to\infty}|\sqrt{\frac{n}{n+1}}|$. I want to use L'Hospitals rule but I'm uncertain how to proceed. Can I use L'H rule on $\frac{n}{n+1}$ or do I use it on $\frac{\sqrt{\frac{n}{n+1}}}{1}$
Keep in mind that $\lim\left|x\right|\left|\sqrt{\frac{n}{n+1}}\right |<1$

Now note that $\sqrt{\frac{n}{n+1}}=\sqrt{\frac{1}{1+\frac{1}{n}} }$.

So what is $\lim\left|\sqrt{\frac{1}{1+\frac{1}{n}}}\right|$?

EDIT: Skeeter beat me to it...

4. It's not apparent to me how
$\sqrt{\frac{n}{n+1}} = \sqrt{\frac{1}{1+\frac{1}{n}}}$

5. Originally Posted by superdude
It's not apparent to me how
$\sqrt{\frac{n}{n+1}} = \sqrt{\frac{1}{1+\frac{1}{n}}}$
Divide the numerator and denominator inside the radical on the left hand side of the equation by n and see what happens

6. Originally Posted by VonNemo19
Divide the numerator and denominator inside the radical on the left hand side of the equation by n and see what happens
does doing that sum how not effect the number? I mean you can't just go dividing things. For example, is this equivalent to multiplying by 1.

7. Originally Posted by superdude
does doing that sum how not effect the number? I mean you can't just go dividing things. For example, is this equivalent to multiplying by 1.

We can always multiply (in your case $\frac{1}{n}$) top and bottom of a fraction by the same value because it does not affect the value of the ratio that is implied.

EG $\frac{ca}{cb}=\frac{a}{b}$

8. got it, thanks