### Select your language

Suggested languages for you:

Americas

Europe

Q4E

Expert-verified
Found in: Page 433

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

### Answers without the blur.

Just sign up for free and you're in.

# In problems 1-6, determine the convergence set of the given power series.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{4}}{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{n}}{\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{\right)}}^{{\mathbf{n}}}$

The set is, $x\in \left[2,4\right]$

See the step by step solution

## Step 1:To Find the Radius of convergence

Use the ratio test to determine the radius of convergence.

$\begin{array}{c}\underset{n\to \infty }{\mathrm{lim}}|\frac{{a}_{n}}{{a}_{n+1}}|=\underset{n\to \infty }{\mathrm{lim}}\left|\frac{\frac{4}{{n}^{2}+2n}}{\frac{4}{{\left(n+1\right)}^{2}+2\left(n+1\right)}}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{{n}^{2}+2n}{{n}^{2}+2n+1+2n+2}×\frac{4}{4}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{{n}^{2}+2n}{{n}^{2}+4n+3}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{{n}^{2}\left(1+\left(2/n\right)\right)}{{n}^{2}\left(1+\left(4/n\right)+\left(3/{n}^{2}\right)\right)}\right|\\ =\underset{n\to \infty }{\mathrm{lim}}\left|\frac{\left(1+\left(2/n\right)\right)}{\left(1+\left(4/n\right)+\left(3/{n}^{2}\right)\right)}\right|\\ =1\end{array}$

The radius of convergence is 1, therefore convergent set for the given power series is .

$|x-3|<1$

## Step 2: Find the set of convergence

To completely identify the convergence set, we have to check whether the boundary points 2 and 4 are included in the set or not.

Checking at $x=2$, by substituting x by 2,

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{4}{{n}^{2}+2n}{\left(x-3\right)}^{n}=\sum _{n=0}^{\infty }\frac{4}{{n}^{2}+2n}{\left(2-3\right)}^{n}\\ =\sum _{n=0}^{\infty }\frac{4}{{n}^{2}+2n}{\left(-1\right)}^{n}\end{array}$

The above series is an alternating harmonic series, which is convergent in nature, thus the point 2 is included in the convergent set.

Similarly, checking at x-4, by substituting x by 4 ,

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{4}{{n}^{2}+2n}{\left(x-3\right)}^{n}=\sum _{n=0}^{\infty }\frac{4}{{n}^{2}+2n}{\left(4-3\right)}^{r}\\ =\sum _{n=0}^{\infty }\frac{4}{{n}^{2}+2n}\end{array}$

Since, ${n}^{2}+2n>{n}^{2}⇒\frac{1}{{n}^{2}+2n}<\frac{1}{{n}^{2}}$and$\sum \frac{1}{{n}^{2}}$is convergent, then, by the comparison test it follows that$\sum \frac{1}{{n}^{2}+2n}$ is also convergent.

The convergent set for the given power series is.$x\in \left[2,4\right]$

### Want to see more solutions like these?

Sign up for free to discover our expert answers

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.