Question

\[\sum_{n=1}^{\infty} {n^2}/{2^n}\]

I'm supposed to use the ratio test to see if this series is absolutely convergent, conditionally convergent, or divergent.

Answer 1

\[{n+1}^{2}/ {2}^{n+1}\] is greater than the original equation

Answer 2

you need the absolute value of the ratio of successive terms though

Answer 3

sorry i meant (n+1)

Answer 4

ohh i need to see if the absolute value of the ratio is greater or less than 1?

Answer 5

\[ \lim_{n \rightarrow \infty} \left ( \frac{(n+1)^2}{2^{n+1}} \over \frac{n^2}{2^{n}}\right ) = \lim_{n \rightarrow \infty} \frac{ \left( \frac{n + 1}{n} \right )^2}{2} = \frac 1 2 < 1\]

Answer 6

\[\frac{(n+1)^2\times 2^n}{2^{n+1}n^2}\]

Answer 7

you get
\[\frac{(n+1)^2}{2n^2}\] take limit as \(n\to \infty\) and get \(\frac{1}{2}\)

Answer 8

makes sense...so this series is absolutely convergent

Answer 9

if it is convergent it is certainly absolutely convergent because all the terms are positive

Answer 10

btw experimentx is of course right, but i find it easier not to write a compound fraction if i can get away with it. i.e if your term \(a_n\) is a fraction multiply by the reciprocal when you take your ratio

Answer 11

thanks so much everyone!!!

Answer 12

whatever you are most comfortable with