Using Ratio Test for a Series with odd and even indexed coefficients

Question

The problem I have is the following:

Suppose the radius of convergence of the series $\sum_{n=0}^\infty a_n z^n$ is equal to $2$. Find the radius of convergence of the series $$\sum_{n=0}^\infty 2^{\sqrt n} (|a_{2n} + |a_{2n+1}|)z^{3n}$$

By the ratio test, I know that $$\lim_{n\to \infty} |\frac{a_{n+1}}{ a_n}| = 1/2 $$

So that the series converges whenever $|z|<1/2$.

I want to compute the radius of convergence for the other series using the ratio test, however, I have doubts on how to set it up, would I be computing:

$$\lim_{n\to \infty} \frac{2^{\sqrt{n+1}} (|a_{2n+2}| +|a_{2n+3} |) }{2^{\sqrt n} (|a_{2n}| +|a_{2n+1}|)} $$

In doing this computation, I found that the limit is $1/4$, so the series converges when $|z|< 4^{1/3}$.

I am not sure if I should be computing a limit superior here being that the power of $z$ is $3n$ instead of $n$. If I do need to use a limit superior, how would I index the ratio of my terms?

Answer 1

I just describe you what should have been done and do not dwell into computing the radius of convergence.

For a power series $\sum\limits_{n=0}^{\infty}a_nz^n$, formally, the radius of convergence $R$ is defined to be $$R = \frac{1}{\lim\sup\limits_{n\to\infty}|a_n|^{\frac{1}{n}}}.\tag{1}$$ Where, conventionally we take $R=0$ if the limit supremum is infinity and we take $R=0$, if the limit supremum is $0$.

We do not rely on ratio test for finding the radius of convergence. Observe that, the radius of convergence $R$ is such a number that the power series $\sum\limits_{n=0}^{\infty}a_nz^n$ converges if $|z|<R$ and diverges if $|z|>R$. The definition in $(1)$ satisfies this and this was arrived at through root test. There was an advantage in root test which says if $\sum\limits_{n}a_n$ is a series, and $\alpha=\lim\sup\limits_{n\to \infty} |a_n|^{\frac{1}{n}}$, then

1. The series converges if $\alpha<1$
2. The series diverges if $\alpha>1$.

Note that for both convergence and divergence, we keep the same quantity, i.e. limit supremum.

While in ratio test, which says if $\lim\sup\limits_{n\to \infty}\frac{a_{n+1}}{a_n}=\alpha$, then the series converges if $\alpha<1$. But it does not tell that the series diverges if $\alpha>1$. For instance, consider the series

$$\frac{1}{2}+1+\frac{1}{8}+\frac{1}{4}+\frac{1}{32}+\frac{1}{16}+\dots$$

Clearly, $\lim\sup\limits_{n\to \infty}\frac{a_{n+1}}{a_n}$ of the above series is $2>1$ but the series converges. Hence, ratio test cannot endow the property which we are looking for in $R$, the radius of convergence.

Also if the radius of convergence is $R$, then you cannot conclude from it that $$\lim\sup\limits_{n\to \infty}\frac{a_{n+1}}{a_n}=\frac{1}{R}$$

Because, for the series

$$\frac{1}{2}+x+\frac{1}{8}x^2+\frac{1}{4}x^3+\dots,$$ the radius of convergence is $2$, while also $\lim\sup\limits_{n\to \infty}\frac{a_{n+1}}{a_n}=2$.

Only under certain circumstances such as: if $a_n$ is positive and if $\lim\limits_{n\to \infty}\frac{a_{n+1}}{a_n}$ exists, then we have in this situation the quantity in $(1)$ equals this limit.

Answer 2

The radius of convergence is equal to the largest $r$ such that $a_nz^n\to 0,$ for $|z|<r.$ If $r=2$ then $|a_n||z|^n\to 0$ for $|z|<2.$ Hence $$|a_{2n}|\,|z|^{3n}=|a_{2n}|\,(|z|^{3/2})^{2n}\to 0, \quad |z|^{3/2}<2$$ Similarly $$|a_{2n+1}|\,|z|^{3n}=|a_{2n+1}|\, (|z|^{3/2})^{2n+1}\,|z|^{-3/2}\to 0, \quad 0<|z|^{3/2}<2$$ Thus the series $$\sum_{n=1}^\infty (|a_{2n}|+|a_{2n+1}|)z^{3n}\quad (*)$$ is convergent if $|z|<2^{2/3}.$ Observe that the reasoning can be reversed, i.e. the convergence of the series $(*)$ implies the convergence of the original series. Hence the radius of convergence of $(*)$ cannot be greater than $2^{2/3},$ i.e. it is equal $2^{2/3}.$ The radius of convergence of $$\sum_{n=1}^\infty 2^{\sqrt{n}}(|a_{2n}|+|a_{2n+1}|)z^{3n}\quad (**)$$ cannot be greater than that of $(*).$ Let $|z|< 2^{2/3},$ i.e. $|z|^3< 4.$ Then $|z|^3<4-2\delta$ for some $\delta>0.$ Denote $t=(4-2\delta)/(4-\delta).$ Then $t<1$ and $$2^{\sqrt{n}}(|a_{2n}|+|a_{2n+1}|)\,|z|^{3n}<2^{\sqrt{n}}(|a_{2n}+|a_{2n+1}|)\,(4-\delta)^nt^n$$ The sequence $(|a_{2n}|+|a_{2n+1}|)(4-\delta)^n$ tends to $0$ (it corresponds to $z=(4-\delta)^{1/3}).$ The sequence $2^{\sqrt{n}}t^n$ tends to $0$ as well by the $n$-root test.

Finally we may conclude that the radius of convergence of the series in OP is indeed equal $2^{2/3}.$