Chapter 8, Problem 33RQ

To determine

The radius of convergence and interval of convergence.

Answer to Problem 33RQ

The radius of convergence is $\frac{1}{2}$ and interval of convergence $\left(\frac{5}{2},\frac{7}{2}\right)$ .

Explanation of Solution

Given:

The series is given as ${\displaystyle \sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2^{\mathrm{n}}{\left(\mathrm{x}-3\right)}^{\mathrm{n}}}{\sqrt{\mathrm{n}+3}}}$ .

Compare the given series with standard form ${\displaystyle \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}}{\left(\mathrm{x}-\mathrm{b}\right)}^{\mathrm{n}}$ .

Here, ${\mathrm{a}}_{\mathrm{n}}=\frac{2^{\mathrm{n}}}{\sqrt{\mathrm{n}+3}}$ and $\mathrm{b}=3$

Calculate the radius of convergence of the given series as,

  $\begin{array}{c}\frac{1}{\mathrm{R}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left|\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}\right|\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left|\frac{2^{\mathrm{n}}\cdot 2}{\sqrt{\mathrm{n}+4}}\times \frac{\sqrt{\mathrm{n}+3}}{2^{\mathrm{n}}}\right|\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left|\frac{2\sqrt{\mathrm{n}}\sqrt{1+\frac{3}{\mathrm{n}}}}{\sqrt{\mathrm{n}}\sqrt{1+\frac{4}{\mathrm{n}}}}\right|\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left|\frac{2\sqrt{1+\frac{3}{\mathrm{n}}}}{\sqrt{1+\frac{4}{\mathrm{n}}}}\right|\end{array}$

Apply the limit of the series.

  $\begin{array}{l}\frac{1}{\mathrm{R}}=\frac{2\sqrt{1+0}}{\sqrt{1+0}}\\ \frac{1}{\mathrm{R}}=\frac{2\sqrt{1}}{\sqrt{1}}\\ \frac{1}{\mathrm{R}}=2\\ \mathrm{R}=\frac{1}{2}\end{array}$

Therefore, the radius of the convergence is $\frac{1}{2}$ .

Calculate the interval of convergence of the given series as,

  $\begin{array}{c}\left|\mathrm{x}-\mathrm{b}\right|<\mathrm{R}\\ \left|\mathrm{x}-3\right|<\frac{1}{2}\\ -\frac{1}{2}<\mathrm{x}-3<\frac{1}{2}\\ \frac{5}{2}<\mathrm{x}-3<\frac{7}{2}\end{array}$

Therefore, the radius of convergence is $\frac{1}{2}$ and interval of convergence $\left(\frac{5}{2},\frac{7}{2}\right)$ .