Question

the radius of convergence of the series x^n/(n*2^n)

Answer 1

I need help with understanding how to solve for the radius of convergence

Answer 2

Ratio test:
\[\lim_{n\to\infty}\left|\frac{x^{n+1}}{(n+1)2^{n+1}}\cdot\frac{n2^n}{x^n}\right|=\lim_{n\to\infty}\left|\frac{x}{2}\cdot\frac{n}{n+1}\right|=\frac{1}{2}|x|\lim_{n\to\infty}\frac{n}{n+1}\]
What are the convergence conditions for the ratio test?

Answer 3

n<1 is converging?

Answer 4

*x not n

Answer 5

That would be right if we were left with just \(|x|\). In general, the limit must be less than 1, otherwise it diverges.

Above, the expression containing \(n\) approaches 1, so we're left with \(\dfrac{|x|}{2}\). If this series only converges if this value is less than 1, then convergence is attained when \(|x|<2\).

Answer 6

okay thanks!! so basically, you get abs(x)/2 and do you just say that abs(x)/2<1 or how do you determine what the number is that is on the right side of the less than sign?

Answer 7

When you do the ratio test, you'll have something that looks like this:
\[\lim_{n\to\infty}\bigg|\cdots\bigg|\]
from you should be able to extract something that looks like this:
\[c|x|\lim_{n\to\infty}\bigg|\cdots\bigg|\]
Evaluate the limit (I'll call it \(L\)), then solve for \(|x|\) in the inequality,
\[c|x|L<1~~\Rightarrow~~|x|<\frac{1}{cL}\]
The radius is given by \(\dfrac{1}{cL}\).

Answer 8

In this case, you had \(L=1\) and \(c=\dfrac{1}{2}\), which makes \(\dfrac{1}{cL}=2\).

Answer 9

okay that makes sense thanks!!

Answer 10

yw