Examples

Radius of convergence

Find the radius of convergence of the series:

(a) ${\displaystyle \sum _{n=0}^{\infty }\frac{x^n}{2^n}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{2n}}{(2n)!}}$

Solution

(a) Ratio of terms:

$$\begin{array}{c}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{{\displaystyle \frac{x^{n+1}}{2^{n+1}}}}{{\displaystyle \frac{x^n}{2^n}}}\right|\\ \\ \gg \gg |\frac{x^{n+1}}{2^{n+1}}\cdot \frac{2^n}{x^n}|\gg \gg \frac{1}{2}|x|=L\end{array}$$

This converges for $L<1$, or $|x|<2$. Therefore $R=2$.

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(b) This power series skips the odd powers. Apply the ratio test to just the even powers:

$$\begin{array}{c}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{{\displaystyle \frac{x^{2n+2}}{(2n+2)!}}}{{\displaystyle \frac{x^{2n}}{(2n)!}}}\right|\\ \\ \gg \gg |\frac{x^{2n+2}}{(2n+2)!}\cdot \frac{(2n)!}{x^{2n}}|\\ \\ \gg \gg \frac{1}{(2n+2)(2n+1)}|x^2|\gg \gg R=\infty \end{array}$$

Interval of convergence

Find the radius and interval of convergence of the following series:

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(x-3)}^n}{n}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-3)}^n{x}^n}{\sqrt{n+1}}}$

Solution

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(x-3)}^n}{n}}$

(1) Apply ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|{\displaystyle \frac{{(x-3)}^{n+1}}{n+1}}\cdot {\displaystyle \frac{n}{{(x-3)}^n}}|\\ \\ \gg \gg \frac{n}{n+1}|x-3|\stackrel{n\to \infty }{⟶}|x-3|=L\end{array}$$

Therefore $R=1$ and thus:

$$\begin{array}{cc}|x-3|<1& ⟹\text{converges}\\ & \\ |x-3|>1& ⟹\text{diverges}\end{array}$$

Preliminary interval: $x\in (\mathrm{2,4})$.

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(2) Check endpoints:

Check endpoint $x=2$:

$$\begin{array}{c}\sum _{n=1}^{\infty }\frac{{(2-3)}^n}{n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}\\ \\ \gg \gg \text{converges by AST}\end{array}$$

Check endpoint $x=4$:

$$\begin{array}{c}\sum _{n=1}^{\infty }\frac{{(4-3)}^n}{n}\gg \gg \sum _{n=1}^{\infty }\frac{1}{n}\\ \\ \gg \gg \text{diverges as }p\text{-series}\end{array}$$

Final interval of convergence: $x\in [\mathrm{2,4})$

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(b) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-3)}^n{x}^n}{\sqrt{n+1}}}$

(1) Apply ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|\frac{{(-3)}^{n+1}{x}^{n+1}}{\sqrt{n+2}}\cdot \frac{\sqrt{n+1}}{{(-3)}^n{x}^n}|\\ \\ \gg \gg \frac{|(-3){(-3)}^n|}{\sqrt{n+2}}\cdot \frac{\sqrt{n+1}}{|{(-3)}^n|}|x|\\ \\ \gg \gg \frac{3\sqrt{n+1}}{\sqrt{n+2}}|x|\stackrel{n\to \infty }{⟶}3|x|=L\end{array}$$

Therefore:

$$\begin{array}{cc}|x|<\frac{1}{3}& ⟹\text{converges}\\ & \\ |x|>\frac{1}{3}& ⟹\text{diverges}\end{array}$$

Preliminary interval: $x\in (-{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}})$

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(2) Check endpoints:

Check endpoint $x=-1/3$:

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{(-3\cdot (-\frac{1}{3}))}^n}{\sqrt{n+1}}\gg \gg \sum _{n=0}^{\infty }\frac{1}{\sqrt{n+1}}\\ \\ \gg \gg \text{diverges by LCT with }{b}_n=1/\sqrt{n}\end{array}$$

Check endpoint $x=+1/3$:

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{(-3\cdot (+\frac{1}{3}))}^n}{\sqrt{n+1}}\gg \gg \sum _{n=0}^{\infty }\frac{{(-1)}^n}{\sqrt{n+1}}\\ \\ \gg \gg \text{converges by AST}\end{array}$$

Final interval of convergence: $x\in (-1/\mathrm{3,1}/3]$

Radius and interval for a few series

Find the radius and interval of convergence of the following series:

(a) ${\displaystyle \sum _{n=0}^{\infty }{x}^n}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }n!x^n}$

Interval of convergence - further examples

Find the interval of convergence of the following series.

(a) ${\displaystyle \sum _{n=0}^{\infty }\frac{n{(x+2)}^n}{3^{n+1}}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(4x+1)}^n}{n}}$

Solution

(a) ${\displaystyle \sum _{n=0}^{\infty }\frac{n{(x+2)}^n}{3^{n+1}}}$

Ratio of terms:

$$\frac{n+1}{3^{n+2}}\cdot \frac{3^{n+1}}{n}|x+2|\gg \gg \frac{n+1}{3n}|x+2|\stackrel{n\to \infty }{⟶}\frac{1}{3}|x+2|=L$$

This converges when $L<1$ or $|x-(-2)|<3$.

Therefore $R=3$ and the preliminary interval is $x\in (-\mathrm{5,1})$.

Check endpoints: ${\displaystyle \sum \frac{n{(-3)}^n}{3^{n+1}}}$ diverges and ${\displaystyle \sum \frac{n{(3)}^n}{3^{n+1}}}$ also diverges.

Final interval is $(-\mathrm{5,1})$.

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(b) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(4x+1)}^n}{n}}$

Ratio of terms:

$$\begin{array}{c}\left|\frac{a_{n+1}}{a_n}\right|\gg \gg \frac{{\displaystyle \frac{1}{n+1}}}{{\displaystyle \frac{1}{n}}}|4x+1|\\ \\ \gg \gg \frac{n}{n+1}|4x+1|\stackrel{n\to \infty }{⟶}|4x+1|=L\end{array}$$

This converges when $L<1$ or:

$$\begin{array}{cc}|4x+1|<& 1⟺|x+1/4|<1/4⟹\text{converges}\\ & \\ |4x+1|>& 1⟺|x+1/4|>1/4⟹\text{diverges}\end{array}$$

Preliminary interval: $x\in (-1/\mathrm{2,0})$

Check endpoints: ${\displaystyle \sum \frac{{(4\cdot \frac{-1}{2}+1)}^n}{n}}$ converges but ${\displaystyle \sum \frac{1}{n}}$ diverges.

Final interval of convergence: $[-1/\mathrm{2,0})$