Question

find the values of x for which the series converges. (enter your answer using interval notation.)
\\(\sum_{n = 0}^{\infty} \frac{(x - 4)^n}{9^n}\\)

find the sum of the series for those values of x.

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1. - / 0.88 points

use the integral test to determine whether the series is convergent or divergent.
\\(\sum_{n = 1}^{\infty} n^2 e^{-n^3}\\)

evaluate the following integral.
\\(\int_{1}^{\infty} x^2 e^{-x^3} dx\\)

Explanation:

Step1: Recognize the series type

The series \(\sum_{n = 0}^{\infty}\frac{(x - 4)^{n}}{9^{n}}=\sum_{n=0}^{\infty}(\frac{x - 4}{9})^{n}\) is a geometric series with \(a = 1\) (when \(n = 0\), \((\frac{x - 4}{9})^{0}=1\)) and \(r=\frac{x - 4}{9}\).

Step2: Apply the geometric series convergence condition

A geometric series \(\sum_{n = 0}^{\infty}ar^{n}\) converges if \(|r|\lt1\). So, we set \(|\frac{x - 4}{9}|\lt1\).

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Step3: Find the sum of the geometric series

The sum of a geometric series \(\sum_{n=0}^{\infty}ar^{n}=\frac{a}{1 - r}\) for \(|r|\lt1\). Here, \(a = 1\) and \(r=\frac{x - 4}{9}\), so the sum \(S=\frac{1}{1-\frac{x - 4}{9}}\).

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Answer:

The interval of convergence is \((-5,13)\) and the sum of the series is \(\frac{9}{13 - x}\)