1. ## Series Problem

Hello, I need help with this series:

The instructions are: Find all the values of x such that the given series would converge. The series is convergent
from , left end included (enter Y or N):
to , right end included (enter Y or N):
Now, I used the ratio test but I got suck with the square root of n, could someone show the steps to solve the series with the ratio test? I know after the values are found you put in each value of x in the series then solve using a test like the limit comparison test to see if it converges so that you know if it is included for the values.
Thanks,
Matt

2. $\displaystyle \lim_{n \to \infty} \left|\frac{10^{n+1}x^{n+1}}{\sqrt{n+1}+2} \cdot \frac{\sqrt{n}+2}{10^n x^n} \right| < 1$

$\displaystyle |10x| \lim_{n \to \infty} \frac{\sqrt{n}+2}{\sqrt{n+1}+2} < 1$

$\displaystyle |10x| \lim_{n \to \infty} \frac{1 + \frac{2}{\sqrt{n}}}{\sqrt{1+\frac{1}{\sqrt{n}}}+\f rac{2}{\sqrt{n}}} < 1$

$\displaystyle |10x| \cdot 1 < 1$

interval of convergence is

$\displaystyle -\frac{1}{10} < x \leq \frac{1}{10}$

3. Thank you! But, what happened to the (-1)^n from the numerator from the original series? Wouldn't it be just -1, how does it go away?\

Edit: Oh wait, is it because (-1)^infinity is indeterminate?

4. absolute value ?

5. Also, what method did you use to find out if the endpoints were included for the interval of convergence?

6. Originally Posted by skeeter
absolute value ?
haha, I forgot it was there. Thanks.

7. Originally Posted by matt3D
Also, what method did you use to find out if the endpoints were included for the interval of convergence?
sub each endpoint for x in the original nth term.

for x = 1/10, you get an alternating series whose nth term goes to 0.

for x = -1/10, the series diverges ... why?

8. Is it because p=1/2 which is less than 1 for -1/10, so that makes it divergent? The p-series is from the integral test right?

9. Well after plugging it in and simplifying, yep:

$\displaystyle \sum_{n = 1}^{\infty} \frac{(-1)^n \cdot 10^n \cdot \displaystyle \frac{1}{(-10)^n}}{\sqrt{n + 2}} = \sum_{n = 1}^{\infty} \frac{(-10)^n \cdot \displaystyle \frac{1}{(-10)^n}}{\sqrt{n+2}} = \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n+2}} = \sum_{n=3}^{\infty} \frac{1}{\sqrt{n}}$

10. Cool, I didn't know you could take the 2 from the denominator and add it to n in the limit to make it 3...