Radius of convergence

**p00ndawg** · April 19th 2009, 04:30 PM

summation from n = 0 -> infinity of:

$\sum\frac{n}{2^n}x^n$

how do you find it? completely lost.

I tried divergence and convergence stuff but it doesnt seem to net me the correct answer.

**skeeter** · April 19th 2009, 04:46 PM

use the ratio test to find the radius of convergence ...

$\lim_{n \to \infty} \left|\frac{(n+1)x^{n+1}}{2^{n+1}} \cdot \frac{2^n}{n x^n} \right| < 1$

$\lim_{n \to \infty} \left|\frac{(n+1)x}{2n}\right| < 1$

$\left|\frac{x}{2}\right| \lim_{n \to \infty} \frac{n+1}{n} < 1$

$\left|\frac{x}{2}\right| \cdot 1 < 1$

$-1 < \frac{x}{2} < 1$

$-2 < x < 2$

**p00ndawg** · April 19th 2009, 04:53 PM

Originally Posted by skeeter

use the ratio test to find the radius of convergence ...

$\lim_{n \to \infty} \left|\frac{(n+1)x^{n+1}}{2^{n+1}} \cdot \frac{2^n}{n x^n} \right| < 1$

$\lim_{n \to \infty} \left|\frac{(n+1)x}{2n}\right| < 1$

$\left|\frac{x}{2}\right| \lim_{n \to \infty} \frac{n+1}{n} < 1$

$\left|\frac{x}{2}\right| \cdot 1 < 1$

$-1 < \frac{x}{2} < 1$

$-2 < x < 2$

ahhhhhh I get it. I wasnt sure how to deal with the x and the n together like that.

Thanks alot, ill try one another now. oh quck question, why you set limit to < 1.?

**mollymcf2009** · April 19th 2009, 04:57 PM

Originally Posted by p00ndawg

ahhhhhh I get it. I wasnt sure how to deal with the x and the n together like that.

Thanks alot, ill try one another now. oh quck question, why you set limit to < 1.?

The ratio test for convergence says that for the series to converge,

$\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|$ must be < 1

**p00ndawg** · April 19th 2009, 06:07 PM

Originally Posted by mollymcf2009

The ratio test for convergence says that for the series to converge,

$\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right|$ must be < 1

oh ok thats what i thought, just wanted to make sure.

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