## anonymous 5 years ago Find the interval of convergence of the sum of (x-2)^n divided by the square root of n.

1. anonymous

The answer is [1,3), how did they get to that?

2. anonymous

I'm doing the problem...

3. anonymous

Do you know how to find the biggest chunk of the interval of convergence?

4. anonymous

That is, I'm asking if it's only the end points you're having trouble with.

5. anonymous

?

6. anonymous

oh sorry, i actually figured out what i did wrong. Thanks tho!

7. anonymous

while you're hear though.... how do you do term by term multiplication in a series? i have to find the first four nonzero of the MacLaurin series sinx *cosx

8. anonymous

Anyway, the series is convergent for x such that$\lim_{n \rightarrow \infty}\ \left| \frac{(x-2)^{n+1}/\sqrt{n+1}}{(x-2)^n/\sqrt{n}} \right|<1$ by the ratio test, and then check for convergence at each of the end points, x=1 and x=3. Convergent for 1 by alternating series test, and non-convergent for 3 by integral test, say.

9. anonymous

You form the Cauchy product.

10. anonymous

does anyone know how i can scan my paper

11. anonymous

$\left( \sum_{}{}a_n \right)\left( \sum_{}{}b_n \right)=\sum_{}{}c_n$where$c_n=\sum_{k=0}^{n}a_kb_{n-k}$

12. anonymous

does anyone know how i can scan my paper

13. anonymous

You need a scanner to scan your paper, mary :)

14. anonymous

tell me steps to scan my paper

15. anonymous

$\sum_{n=0}^{\infty}c_n=\sum_{n=0}^{\infty}\left( \sum_{k=0}^{n}a_kb_{n-k} \right)$

16. anonymous

Well, it depends on your software and machinery. They're not all the same. You have to make sure your scanner is connected to your computer, either through a cable or wireless, and use the appropriate software.