Question

Find the interval of convergence of the sum of (x-2)^n divided by the square root of n.

Answer 1

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

Answer 2

I'm doing the problem...

Answer 3

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

Answer 4

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

Answer 5

?

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

You form the Cauchy product.

Answer 10

does anyone know how i can scan my paper

Answer 11

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

Answer 12

does anyone know how i can scan my paper

Answer 13

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

Answer 14

tell me steps to scan my paper

Answer 15

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

Answer 16

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.