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Expert-verified Found in: Page 421 ### 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 # In problems 1-6, determine the convergence set of the given power series. $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{3}}{{\mathbf{n}}^{\mathbf{3}}}{\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\right)}}^{{\mathbf{n}}}$

The set is, $x\in \left[1,3\right]$

See the step by step solution

## Step 1:To Find the Radius of convergence

Use the ratio test to determine the radius of convergence.

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

The radius of convergence is 1, therefore convergent set for the given power series is.$|x-2|<1$

## Step 2: Find the set of convergence

To completely identify the convergence set, we have to check whether the boundary points 1 and 3 are included in the set or not.

Checking at ,$x=1$ by substituting the value of x by 1,

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

The above series is an alternating power series with$p=3>1$ , which is convergent, thus the point 1 is included in the convergent set.

Similarly, checking at $x=3$, by substituting the value of x by 3,

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

The above series is a power series with $p=3>1$, which is convergent, thus the point 3 is included in the convergent set.

The convergent set for the given power series is.$x\in \left[1,3\right]$ ### Want to see more solutions like these? 