# Let $a_n$ be a sequence of nonnegative numbers such that $a_n \rightarrow b$. Then, $\lim_{n \rightarrow \infty} \frac{(a_{n+1})^{n+1}}{(a_n)^n}$ =?

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$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n} = ||f||_\infty$$

• As $n\to \infty$ , $\epsilon \to 0$. Jun 27, 2021 at 13:42
• Isn't it the other way around? as $\epsilon \rightarrow 0$, $n \rightarrow \infty$. Jun 27, 2021 at 13:43
• @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? Jun 27, 2021 at 13:51
• @paulpogba not exactly. $a_n$ approaches $b$ as $n\to \infty$ and as you are writing it as $b\pm \epsilon$ then $\epsilon \to 0$ Jun 27, 2021 at 14:16
• @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? Jun 27, 2021 at 14:51

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.

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 $$+\infty$$.

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 $$1^{\infty}$$.

• Thank you. This is what I was wondering about. Jun 27, 2021 at 16:17
• 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 $\lim a_n=\alpha$. Jun 27, 2021 at 17:11
• 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 $L=0$).
– AGH
Jun 28, 2021 at 15:07