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Question: Does $\limsup_{n \to \infty }\left | \frac{a_{n+1}}{a_{n}} \right |=\limsup_{n \to \infty }\left | a_n \right |^{1/{n}}$?

I guess the identity is true, given $\limsup_{n \to \infty }\left | \frac{a_{n+1}}{a_{n}} \right |\neq 0$, and here is my "proof", is it correct?

Proof:

Suppose all $a_n$ are non-zero, and $s=\limsup_{n \to \infty }\left | \frac{a_{n+1}}{a_{n}} \right | \neq 0$,

For any $\varepsilon >0$, there exists $N\in \mathbb{N}$ such that $n>N$ implies: $$s-\varepsilon < \underset{m\geq n}{\text{sup}}\left | \frac{a_{m+1}}{a_{m}} \right |<s+\varepsilon$$ $$\Rightarrow s-\varepsilon < \left | \frac{a_{n+1}}{a_{n}} \right |<s+\varepsilon$$ for all $n>N$. $$\Rightarrow (s-\varepsilon)|a_{n}| < |a_{n+1}|<(s+\varepsilon)|a_{n}|$$ $$\Rightarrow (s-\varepsilon)^{n-N}|a_{N+1}| < |a_{n+1}|<(s+\varepsilon)^{n-N}|a_{N+1}|$$ where we may set $0<\varepsilon <s$, where $s \neq0$. So, for all $n>N$, we have: $$\Rightarrow (s-\varepsilon)^{1-\frac{N+1}{n+1}}|a_{N+1}|^{1/{n+1}} < |a_{n+1}|^{1/{n+1}}<(s+\varepsilon)^{1-\frac{N+1}{n+1}}|a_{N+1}|^{1/{n+1}}$$ So, for $n>N$: $$ (s-\varepsilon)^{1-\frac{N+1}{n+1}}|a_{N+1}|^{1/{n+1}} \leq \underset{m\geq n}{\text{sup}}|a_{m+1}|^{1/{m+1}}\leq (s+\varepsilon)^{1-\frac{N+1}{n+1}}|a_{N+1}|^{1/{n+1}}$$ Now, $$(s+\varepsilon)^{1-\frac{N+1}{n+1}}|a_{N+1}|^{1/{n+1}}\underset{n \to \infty }{\rightarrow}(s+\varepsilon)(s+\varepsilon)^0(1)=s+\varepsilon$$ and $$(s-\varepsilon)^{1-\frac{N+1}{n+1}}|a_{N+1}|^{1/{n+1}}\underset{n \to \infty }{\rightarrow}(s-\varepsilon)(s-\varepsilon)^0(1)=s-\varepsilon$$

By comparison test, $$s-\varepsilon \leq \limsup_{n \to \infty }\left | a_n \right |^{1/{n}} \leq s+\varepsilon $$ for arbitrarily small $\varepsilon$.

So, $$s=\limsup_{n \to \infty }\left | a_n \right |^{1/{n}}$$ and we are done.

Any help will be appreciated!

GHG
  • 109

1 Answers1

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The two lim sups are not equal. The flaw in your proof is that the assertion $$s-\varepsilon < \underset{m\geq n}{\text{sup}}\left | \frac{a_{m+1}}{a_{m}} \right |<s+\varepsilon \qquad\forall n>N$$ does not imply $$ s-\varepsilon < \left | \frac{a_{n+1}}{a_{n}} \right |<s+\varepsilon \qquad\forall n>N.$$ A counterexample is $$a_n=\begin{cases} 1&\text{$n$ odd}\\2&\text{$n$ even}\end{cases}.$$ With this sequence, $\sup_{m\ge n}|a_{m+1}/a_m|=2$ while the ratio $|a_{n+1}/a_n|$ oscillates between $\frac12$ and $2$. (And $\limsup |a_{n+1}/a_n|=2$ while $\limsup |a_n|^{1/n}=1$.)

grand_chat
  • 38,951
  • If I replace the flaw by:

    Suppose $a_n \neq 0$, and $s=\lim\left | \frac{a_{n+1}}{a_n} \right | \neq 0$. For any $\varepsilon >0$, there exists $N\in \mathbb{N}$ such that $n>N$ $\Rightarrow s-\varepsilon < \left | \frac{a_{n+1}}{a_{n}} \right |<s+\varepsilon \Rightarrow$ for all $n>N$, $s-\varepsilon \leq \underset{m\geq n}{\text{sup}}\left | \frac{a_{m+1}}{a_{m}} \right |\leq s+\varepsilon$.

    Then the original article can be corrected into a proof of the new statement $\lim \left | \frac{a_{n+1}}{a_n} \right |=\limsup \left | {a_n} \right |^{1/n}$, right?

     – GHG Aug 31 '22 at 16:08
  • But the new statement $\lim |a_{n+1}/a_n|=\limsup |a_n|^{1/n}$ is not true; the counterexample shows that the LHS limit need not exist. – grand_chat Aug 31 '22 at 17:21
  • But what if $\lim|a_{n+1}/a_n|$ does exist? Then the statement would be true right? If this limit exists, then the radius of convergence of the power series $\sum a_nx^n$ has the radius of convergence $1/\lim|a_{n+1}/a_n|$, which should be consistent with the Cauchy-Hadamard formula. – GHG Aug 31 '22 at 21:29
  • Do you mean only the following is true: If $\lim|a_{n+1}/a_n|$ exists, then it must be equal to $\limsup|a_n|^{1/n}$. – GHG Aug 31 '22 at 21:36
  • Yes. See https://math.stackexchange.com/q/814500/215011 and https://math.stackexchange.com/q/69386/215011 – grand_chat Sep 01 '22 at 16:24