Question

power series 2 lecture participation: problem 1
(4 points)
find the first five non - zero terms of power series representation centered at ( x = 0 ) for the function below.
( f(x)=\frac{x^{3}}{1 + 5x} )
answer: ( f(x)=1/6+8/11+27/16+64/21+125/26+cdots )
what is the interval of convergence?
answer (in interval notation):
preview my answers submit answers
you have attempted this problem 0 times.
you have unlimited attempts remaining.
page generated october 15, 2025, 8:33:09 pm edt
webwork © 1996 - 2025 | theme: math4 | ww_version: 2.20 | pg_version 2.20
the webwork project

Explanation:

Step1: Recall the geometric series formula

The geometric series is \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\), for \(|r|\lt1\).
We rewrite \(\frac{1}{1+5x}\) as \(\frac{1}{1-(- 5x)}\). So, \(\frac{1}{1 + 5x}=\sum_{n=0}^{\infty}(-5x)^{n}=\sum_{n = 0}^{\infty}(-1)^{n}5^{n}x^{n}\), for \(| - 5x|\lt1\) (i.e., \(|x|\lt\frac{1}{5}\)).

Step2: Multiply by \(x^{3}\)

Since \(f(x)=\frac{x^{3}}{1 + 5x}\), we multiply the power - series of \(\frac{1}{1 + 5x}\) by \(x^{3}\).
\(f(x)=x^{3}\sum_{n = 0}^{\infty}(-1)^{n}5^{n}x^{n}=\sum_{n = 0}^{\infty}(-1)^{n}5^{n}x^{n + 3}\).

Step3: Find the first five non - zero terms

When \(n = 0\): \((-1)^{0}5^{0}x^{0+3}=x^{3}\).
When \(n = 1\): \((-1)^{1}5^{1}x^{1 + 3}=-5x^{4}\).
When \(n = 2\): \((-1)^{2}5^{2}x^{2+3}=25x^{5}\).
When \(n = 3\): \((-1)^{3}5^{3}x^{3 + 3}=-125x^{6}\).
When \(n = 4\): \((-1)^{4}5^{4}x^{4+3}=625x^{7}\).

Step4: Find the interval of convergence

We know that for the geometric series \(\sum_{n = 0}^{\infty}r^{n}\) with \(r=-5x\), the series converges when \(|r|\lt1\).
Set \(| - 5x|\lt1\). Solving for \(x\), we get \(|x|\lt\frac{1}{5}\).
We check the endpoints:

• When \(x=\frac{1}{5}\), the series becomes \(\sum_{n = 0}^{\infty}(-1)^{n}5^{n}(\frac{1}{5})^{n+3}=\sum_{n = 0}^{\infty}(-1)^{n}\frac{1}{5^{3}}\). This is an alternating series, but \(\lim_{n

ightarrow\infty}a_{n}=\frac{(-1)^{n}}{125}\) does not tend to \(0\) (in a non - trivial sense for convergence of series; the terms do not approach \(0\) as \(n\) gets large in a way that satisfies the conditions for series convergence).

• When \(x =-\frac{1}{5}\), the series becomes \(\sum_{n = 0}^{\infty}(-1)^{n}5^{n}(-\frac{1}{5})^{n+3}=\sum_{n = 0}^{\infty}\frac{-1}{125}\). The terms of the series do not approach \(0\) as \(n

ightarrow\infty\).

Answer:

The first five non - zero terms of \(f(x)\) are \(x^{3}-5x^{4}+25x^{5}-125x^{6}+625x^{7}+\cdots\)
The interval of convergence is \((-\frac{1}{5},\frac{1}{5})\)