find-the-radius-of-convergence-and-interval-of-convergence-of-the-series-sum_n-1infty-2nn2xn

Question

Find the radius of convergence and interval of convergence of the series.

$ \sum_{n = 1}^{\infty} 2^nn^2x^n $

Gabriel Rhodes · Accepted Answer

for the radius of convergence. We take the limit as n goes to infinity Absolute value of a n over a n plus one. And we want for this to be less than one. The ends here is this to to the end times and squared. Okay, so up top, we&#x27;re gonna have to the end and squared down in the denominator We have to do the in plus one times in plus one squared. So to the end about about to the M plus one is one half and then n squared over in plus one square it is, and over in plus one squared limit as n goes to infinity of n over in plus one is one this turns into one squared so one half times one squared would just get one half So one half is the radius of convergence. So this is our are for the interval of convergence. We need to figure out whether or not we include minus one half and whether or not we include one half case, we need to check to see whether or not we get convergence for those values. Okay, so this was the sum that we had we had to the end. I&#x27;m in squared times X to the end, and we were going from an equals one to infinity. So one two check are in points. So we needed the check Well, minus one half. So plug in minus one half here and see whether or not we get convergence. So when you plug in minus one half for acts, we get minus one half to the end. So if we rewrite that a little bit, we get minus one to the end times one half to the end. So to the end, one half to the end cancel. And this should be clear that this is not going to converge because the terms do not go to zero. So that is going to diverge. And similarly, when exes one half the terms that we get are going to be two to the end and squared one half to the end. So the terms that were going to be summing up is just in squared, which clearly do not go to zero. So we can&#x27;t possibly have convergence, So that&#x27;s also went to diverge. So both of the end points we have to toss out so the interval of convergence. We throw out both minus one half and one half and just leave it as this open interval here.

Problem

Find the radius of convergence and interval of co…

Problem 10 Medium Difficulty

Find the radius of convergence and interval of convergence of the series.

$ \sum_{n = 1}^{\infty} 2^nn^2x^n $

Answer

Interval of convergence is $\left(-\frac{1}{2}, \frac{1}{2}\right)$
Radius of Convergence is $\frac{1}{2}$

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Kristen K.

Samuel H.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 11

Video Transcript

Calculus: Early Transcendentals

Grace H.

Catherine R.

Kristen K.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

Interval of convergence is $\left(-\frac{1}{2}, \frac{1}{2}\right)$Radius of Convergence is $\frac{1}{2}$

Calculus: Early Transcendentals

Grace H.

Catherine R.

Kristen K.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

Grace H.

Catherine R.

Kristen K.

Samuel H.

Interval of convergence is $\left(-\frac{1}{2}, \frac{1}{2}\right)$
Radius of Convergence is $\frac{1}{2}$