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I've got a start on the question I've written below. I'm hoping for some help to finish it off.

Suppose that the power series $\sum_{n=0}^{\infty}c_n x^n$ has a radius of convergence $R \in (0, \infty)$. Find the radii of convergence of the power series $\sum_{n=0}^{\infty}c_n x^{2n}$ and $\sum_{n=0}^{\infty}c_n x^{n^2}$.

From Hadamard's Theorem I know that the radius of convergence for $\sum_{n=0}^{\infty}c_n x^n$ is $R=\frac{1}{\alpha}$, where

$$\alpha = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}}.$$

Now, applying the Root Test to $\sum_{n=0}^{\infty}c_n x^{2n}$ gives

$$\limsup |a_nx^{2n}|^{\frac{1}{n}}=x^2 \cdot \limsup |a_n|^{\frac{1}{n}}=x^2 \alpha$$

which gives a radius of convergence $R_1 = \frac{1}{\sqrt{\alpha}}$. Now for the second power series. My first thought was to take

$$\limsup |a_nx^{n^2}|^{\frac{1}{n^2}}=|x| \cdot \limsup |a_n|^{\frac{1}{n^2}}$$

but then I'm stuck. I was trying to write the radius of convergence once again in terms of $\alpha$. Any input appreciated and thanks a bunch.

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5 Answers

$$ \limsup_{n\rightarrow\infty} |c_n|^{\frac{1}{n}}=\alpha <\infty $$ gives that

there exists $N\geq 1$ such that if $n>N$ then $|c_n|^{\frac{1}{n}}< \alpha+1$.

Then $|c_n|^{\frac{1}{n^2}}< (\alpha+ 1)^{\frac{1}{n}}$ for all $n>N$.

It follows that $\limsup_{n\rightarrow\infty} |c_n|^{\frac{1}{n^2}}\leq 1$.

Also, there is a subsequence $\{n_k\}$ such that $|c_{n_k}|^{\frac{1}{n_k}}$ converges to $\alpha$. For that subsequence, we have $$ |c_{n_k}|^{\frac{1}{n_k^2}} \rightarrow 1 $$ as $k\rightarrow\infty$.

This implies that $\limsup_{n\rightarrow\infty} |c_n|^{\frac{1}{n^2}}=1$. Therefore, the radius of convergence of $\sum c_n x^{n^2}$ is 1.

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Express $|a_n|^{\frac1{n^2}}$ as $\left(|a_n|^{\frac1{n}}\right)^\frac1n$. Then show that $\limsup\left(|a_n|^{\frac1{n}}\right)^\frac1n=1.$
In general if $(b_n)_{n\in\mathbb N}$ is a positive sequence such that $\limsup b_n \in (0,+\infty)$ then $\limsup\left(b_n\right)^\frac1n=1.$

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The radius of convergence of the series $\sum\limits_na_nx^n$ is the only $R_a$ in $[0,\infty]$ such that $|a_n|s^n\to0$ for every $s\lt R_a$ and $|a_n|s^n\not\to0$ for every $s\gt R_a$.

If $a_{2n}=c_n$ and $a_{2n+1}=0$, then for every $s\lt\sqrt{R_c}$, $|a_n|s^n\to0$ and for every $s\gt\sqrt{R_c}$, $|a_{2n}|s^{2n}=|c_n|(s^2)^n\not\to0$, thus $R_a=\sqrt{R_c}$.

If $a_{n^2}=c_n$ and $a_k=0$ for every $k$ not a square, then for every $s\lt1$, $|a_{n}|s^{n}\to0$ and for every $s\gt1$, $|a_{n^2}|s^{n^2}=|c_n|s^{n^2}\not\to0$ (this last part uses the fact that $R_c$ is finite), thus $R_a=1$.

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The first can be done without recourse to $\limsup$:

If $R$ is the radius of convergence of $\sum_{n=0}^{\infty}c_n x^n$, then the series converges absolutely if $|x|<R$ and diverges if $|x| > R$.

Hence $\sum_{n=0}^{\infty}c_n x^{2n}= \sum_{n=0}^{\infty}c_n (x^2)^n$ converges absolutely if $|x^2| < R$, and diverges if $|x^2| >R$. Hence the radius of convergence must be $\sqrt{R}$.

For the second, note that if we write the power series as $\sum_{n=0}^{\infty}a_n x^n$, then $a_n = 0$ if $n$ is not a square, and $a_n =c_k$ if $n=k^2$. Hence the radius of convergence is $\frac{1}{\rho} = \limsup_n \sqrt[n^2]{|c_n|} $.

We have $\frac{1}{R} = \limsup_n \sqrt[n]{|c_n|}$, hence for all $\epsilon>0$, there exists a subsequence $n_k$ such that $\frac{1}{R} - \epsilon \le \sqrt[n_k]{|c_{n_k}|}$, and for all $\epsilon>0$, there exists $n$ such that $\frac{1}{R} + \epsilon \ge \sqrt[k]{|c_{k}|}$ for all $k \ge n$.

Hence we have $\sqrt[n_k]{\frac{1}{R} - \epsilon} \le \sqrt[n_k^2]{|c_{n_k}|}$, and taking $\limsup$s we have $1 \le \frac{1}{\rho}$. Similarly, the other inequality gives $1 \ge \frac{1}{\rho}$. Hence $\rho = 1$.

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The question assumes $R\ne0,\infty$. Just for completeness let's see what happens if $R=0$ or $R=\infty$: then the radius of convergence of $\sum c_nx^{n^2}$ is undetermined. Consider the series $$ \sum a^{n^2}x^n,\quad a>0. $$ If $0<a<1$ then the radius of convergence is $R=\infty$, while if $a>1$ then $R=0$. But the radius of convergence of $\sum a^{n^2}x^{n^2}$ is $1/a$.

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