Tag: Radius of Convergence

Determine the radius of convergence of ∑ (sin (an))zⁿ

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=0}^{\infty} (\sin (an))z^n \qquad a> 0.$

Test for convergence at the boundary points if $r$ is finite.

If $|z| \geq 1$ then $\lim_{n \to \infty} (\sin (an))z^n \neq 0$ unless $a = k \pi$ . Thus, the series does not converge for $a \neq k \pi$ if $|z| \geq 1$ .

If $|z| < 1$ then we apply Dirichlet’s test where

$\sum_{n=1}^N | \sin (an) |$

is bounded for any $N$ (i.e., the partial sums are bounded) and the $z^n$ terms are monotonically decreasing with $\lim_{n \to \infty} z^n = 0$ . Hence, the series converges for $|z| < 1$ , which implies $r = 1$ if $a \neq k \pi$ .

If $a = k \pi$ then $\sin (an) = 0$ for all $n$ so the series converges for all $z$ which implies $r = +\infty$ .

Determine the radius of convergence of ∑ (1 + 1/n)^n² zⁿ

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} \left( 1 + \frac{1}{n} \right)^{n^2} z^n.$

Test for convergence at the boundary points if $r$ is finite.

We have

$a_n = \left( 1 + \frac{1}{n} \right)^{n^2} z^n \quad \implies \quad a_n^{\frac{1}{n}} = \left( 1 + \frac{1}{n} \right)^n z.$

Therefore,

$\begin{align*} \lim_{n \to \infty} a_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n z \\[9pt] &= e\cdot z. \end{align*}$

Hence, $r = \frac{1}{e}$ and the series converges for $|z| < \frac{1}{e}$ .

Determine the radius of convergence of ∑ ((135…(2n-1)) / (246…(2n)))³ z

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} \right)^3 z^n.$

Test for convergence at the boundary points if $r$ is finite.

We have

$a_n = \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} \right)^3 z^n.$

Then applying the ratio test we have

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n+2)} \right)^3 \cdot z^{n+1} \cdot \left( \frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)}{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)} \right)^3 \left( \frac{1}{z^n} \right) \\[9pt] &= \lim_{n \to \infty} \left( \frac{(2n+1)^3}{(2n+2)^3} \right) z \\ &= z. \end{align*}$

Hence, the radius of convergence is 1, and the series converges for $|z| < 1$ .

Determine the radius of convergence of ∑ (3^{n^1/2} zⁿ / n)

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} \frac{3^{\sqrt{n}} z^n}{n}.$

Test for convergence at the boundary points if $r$ is finite.

Let $a_n = \frac{3^{\sqrt{n}} z^n}{n}$ . Then,

$\begin{align*} \lim_{n \to \infty} a_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( \frac{3^{\sqrt{n}} z^n}{n} \right)^{\frac{1}{n}} \\[9pt] &= \lim_{n \to \infty} \frac{3^{\frac{1}{\sqrt{n}}} z}{n^{\frac{1}{n}}} \\[9pt] &= z \end{align*}$

Thus, the radius of convergence is $r = 1$ , and the series converges for $z$ such that $|z| < 1$ .

Determine the radius of convergence of ∑ ((n!)² / (2n!)) zⁿ

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n!)} z^n.$

Test for convergence at the boundary points if $r$ is finite.

Let $a_n = \frac{(n!)^2 z^n}{(2n)!}$ . Then we have

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)!^2 z^{n+1}}{(2n+2)!} \right) \left( \frac{(2n)!}{(n!)^2 z^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{ (n+1)^2}{(2n+1)(2n+2)} \right) z\\ &= \frac{z}{4}. \end{align*}$

Thus, the radius of convergence is $r = 4$ . This diverges on the boundary points since if $|z| = 4$ then the terms do not go to 0.

Determine the radius of convergence of ∑ a^n² zⁿ

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=0}^{\infty} a^{n^2} z^n, \qquad 0 < a < 1.$

Test for convergence at the boundary points if $r$ is finite.

Let $b_n = \left( a^n z \right)^n$ . Then,

$\begin{align*} \lim_{n \to \infty} b_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( \left( a^n z \right)^n \right)^{\frac{1}{n}} \\[9pt] &= \lim_{n \to \infty} a^n z \\[9pt] &= 0 \end{align*}$

for all $0 < a < 1$ for any $z$ . Therefore, the radius of convergence is $r = + \infty$ .

Determine the radius of convergence of ∑ (-1)ⁿ(z+1)ⁿ / (n² + 1)

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=0}^{\infty} \frac{(-1)^n (z+1)^n}{n^2 + 1}.$

Test for convergence at the boundary points if $r$ is finite.

Let $a_n = \frac{(z+1)^n}{n^2+1}$ . Then we have

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{ (z+1)^{n+1}}{(n+1)^2 + 1} \right) \left( \frac{n^2+1}{(z+1)^n} \right) \\[9pt] &= \lim_{n \to \infty} (z+1) \left( \frac{n^2+1}{n^2+2n+2} \right) \\[9pt] &= z+1. \end{align*}$

Thus, the series converges for $z$ such that $|z+1| < 1$ ; hence, $r = 1$ . Furthermore, the series converges on the boundary $|z+1| = 1$ by comparison with the series $\sum \frac{1}{n^2}$ .

Determine the radius of convergence of ∑ (n! zⁿ) / nⁿ

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} \frac{n! z^n}{n^n}.$

Test for convergence at the boundary points if $r$ is finite.

Let $a_n = \frac{n! z^n}{n^n}$ . Then,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)! z^{n+1}}{(n+1)^{n+1}} \right) \left( \frac{n^n}{n! z^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{z n^n (n+1)}{(n+1)^{n+1}} \\[9pt] &= \lim_{n \to \infty} \left( \frac{zn^{n+1}}{(n+1)^{n+1}} + \frac{zn^n}{(n+1)^{n+1}} \right) \\[9pt] &= \lim_{n \to \infty} \left( \frac{n}{n+1} \cdot \left( \frac{n}{n+1} \right)^n z \right) \\[9pt] &= 1 \cdot \frac{1}{e} \cdot z \\[9pt] &= \frac{z}{e}. \end{align*}$

Thus, the series converges if $\left| \frac{z}{e} \right| < 1$ which implies $|z| < e$ or the radius of convergence is $r = e$ . It diverges on the boundary points since the terms do not go to 0.

Determine the radius of convergence of (1 – (-2)ⁿ)zⁿ

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} (1 - (-2)^n)z^n.$

Test for convergence at the boundary points if $r$ is finite.

We have

$\begin{align*} \sum_{n=1}^{\infty} \left| (1 - (-2)^n)z^n \right| &= \sum_{n=1}^{\infty} \left|2^{n+1} \right| \left|z^n\right| \\[9pt] &= 2 \sum_{n=1}^{\infty} \left| 2z \right|^n. \end{align*}$

This is a geometric series which converges if and only if $|2z| < 1$ which implies $|z| < \frac{1}{2}$ . Hence, $r = \frac{1}{2}$ , and this converges at none of the boundary points.

Determine the radius of convergence of ∑ (-1)ⁿ2²ⁿz²ⁿ / 2n

by RoRi

March 31, 2016

Determine the radius of convergence $r$ of the power series:

$\sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} z^{2n}}{2n}.$

Test for convergence at the boundary points if $r$ is finite.

This is an alternating series, and we have

$\left| \frac{(-1)^n 2^{2n} z^{2n}}{2n} \right| = \frac{|2z|^{2n}}{2n}.$

If $|2z| > 1$ then the terms are not going to 0 so the series is divergent. If $|2z| < 1$ then it is monotonically decreasing, so we know (by the Leibniz rule) that the series converges if and only if

$\lim_{n \to \infty} \frac{|2z|^{2n}}{2n} = 0 \quad \iff \quad |z| \leq \frac{1}{2}.$

Therefore, the radius of convergence is $r = \frac{1}{2}$ , and the series converges at every boundary point $|z| = \frac{1}{2}$ .