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As the title says, I would like to evaluate the following limit given that $a_n \rightarrow b < \infty$: $\lim_{n \rightarrow \infty} \frac{(a_{n+1})^{n+1}}{(a_n)^n}$ (I am assuming that the limit would evaluate to $b$.)

If $n$ is taken sufficiently large enough, then $|a_n - b|< \epsilon$. Hence, $a_n \in (b-\epsilon, b+\epsilon)$. Hence,

\begin{align*} \frac{(a_{n+1})^{n+1}}{(a_n)^n} &= a_{n+1}\frac{(a_{n+1})^n}{(a_n)^n} \\ &= a_{n+1} \frac{(b+\epsilon)^n}{(b-\epsilon)^n} \\ &= a_{n+1} \frac{b^n + \binom{n}{1} b^{n-1}\epsilon + \cdots + \binom{n}{n} \epsilon^n}{b^n - \binom{n}{1} b^{n-1}\epsilon + \cdots + (-1)^n\binom{n}{n} \epsilon^n} \end{align*}

I am tempted to say that since $b$ term dominates $\epsilon$ terms, the fraction is really equivalent to $\frac{b^n}{b^n}$, but how do I rigorously prove it?


EDIT:

Note that: $$\infty = \lim_{m \rightarrow \infty}\lim_{n \rightarrow \infty} \left(\frac{b+\frac{1}{m}}{b-\frac{1}{m}}\right)^n \neq \lim_{n \rightarrow \infty}\lim_{m \rightarrow \infty} \left(\frac{b+\frac{1}{m}}{b-\frac{1}{m}}\right)^n = 1$$

Is what I have written above a correct approach to the problem?


EDIT: Original motivation of the problem:

Suppose $\mu$ is a positive measure on $X, \mu(X) < \infty, f \in L^\infty(\mu),||f||_\infty > 0, \text{and}$ $$a_n = \int_X|f|^n\,d\mu~~~~~(n=1,2,3,...).$$ prove that$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n} = ||f||_\infty$$

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  • $\begingroup$ As $n\to \infty $ , $\epsilon \to 0$. $\endgroup$ Jun 27, 2021 at 13:42
  • $\begingroup$ Isn't it the other way around? as $\epsilon \rightarrow 0$, $n \rightarrow \infty$. $\endgroup$
    – James C
    Jun 27, 2021 at 13:43
  • $\begingroup$ @LeeMosher I have been thinking for a while, but not sure how to think. Can you provide me a direction how to approach this problem? $\endgroup$
    – James C
    Jun 27, 2021 at 13:51
  • $\begingroup$ @paulpogba not exactly. $a_n$ approaches $b$ as $n\to \infty$ and as you are writing it as $b\pm \epsilon$ then $\epsilon \to 0$ $\endgroup$ Jun 27, 2021 at 14:16
  • $\begingroup$ @LeeMosher Now, the answer is found. However, I am interested in hearing your approach to the problem. What point of convergence property are you highlighting at? $\endgroup$
    – James C
    Jun 27, 2021 at 14:51

3 Answers 3

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Let $a_n:=1+\frac 1{2n}(1-(-1)^n)$, i.e. $a_{2n}=1$ and $a_{2n+1}=1+\frac 1n$. Then the quantity of interest is $A_n:=\frac{a_{n+1}^{n+1}}{a_{n}^{n}}$ has two subsequences converging to $e$ and $1/e$, respectively for even and odd values of $n$. Thus $\lim_{n\to+\infty} A_n$ may not exist.

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Similar to Giulio. Let $$ a_n = 1 + \frac{(-1)^n}{\sqrt{n}} $$ Then $a_{n+1}^{n+1}/a_n^{n}$ has subsequences converging to $0$ and to $+\infty$.

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Here is an approach revealing why the limit need not always exist and how we can evaluate the limit given the sequence $a_n$.

$$\frac{(a_{n+1})^{n+1}}{(a_n)^n} =a_{n+1}\left[1+\left(\frac{a_{n+1}}{a_n}-1\right)\right]^n $$ $$=a_{n+1}\left[1+\left(\frac{a_{n+1}}{a_n}-1\right)\right]^{\frac{1}{\frac{a_{n+1}}{a_n}-1}(\frac{a_{n+1}}{a_n}-1)(n)}$$

Therefore, $\frac{(a_{n+1})^{n+1}}{(a_n)^n}$ approaches $$b\exp\left(\frac{L}{b}\right)$$ where $$L=\lim \limits_{n \to \infty} {(a_{n+1}-a_n)n}$$ assuming it exists. Clearly we cannot decide on $L$ as it can be $0$ or $\infty$ or may not exist.

Further, you cannot conclude out of temptation (referring your question) because the limit in the fraction is of form $1^{\infty}$.

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  • $\begingroup$ Thank you. This is what I was wondering about. $\endgroup$
    – James C
    Jun 27, 2021 at 16:17
  • $\begingroup$ Not that I followed that, but $b\exp(L/b)$ can't be right unless $L=0$. Because the Cesaro-Stolz theorem shows that if $\lim a_{n+1}^{n+1}/a_n^n=\alpha$ then $\lim a_n=\alpha$. $\endgroup$ Jun 27, 2021 at 17:11
  • $\begingroup$ Yes you are right. If $\frac{(a_{n+1})^{n+1}}{(a_n)^n}$ has a limit, then $L=0$ as $L$ can take value either infinity or zero ($L$ has to exist as the whole limit exists and so $L=0$). $\endgroup$
    – AGH
    Jun 28, 2021 at 15:07

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