Show that $$\lim_{n \to \infty} \sup_{k \ge n} \left(\frac{1+a_{k+1}}{a_k}\right)^k \geq e$$ for any sequence $\{a_k\}$ with positive terms, and that this estimate cannot be improved.
Let $$s_k = \left(\frac{1+a_{k+1}}{a_k}\right)^k $$

Then $s_k \geq 0$ as otherwise $a_{k+1} < -1$. Also, $\sup_{k \geq n}\{s_k\}$ is nonincreasing on $n$.
Assume $\lim\limits_{n \to \infty}\sup_{k \geq n}\{s_k\} = l < e$. For $0 < \varepsilon< e-l $, choose $N$ such that $l \leq \sup_{k \geq n}\{s_k\} < e - \varepsilon$ for all $n > N$.

We know that $$\lim\limits_{n \to \infty} \left(\frac{n+1}{n}\right)^n = e$$ and that $r_n = \left(\frac{n+1}{n}\right)^n$ is increasing. Choose $N'$ such that $e- \varepsilon < r_n < e$ for all $n > N'$.

Then for $M =\max\{N,N'\}$, $ \sup_{k \geq n}\{s_k\} <r_n$ for all $n>M$. Furthermore, $s_n \leq \sup_{k \geq n}\{s_k\}$, so $s_n < r_n$.

Here I wanted to get some sort of contradiction but I'm not really sure how. I know that $a_n$ cannot have a limit $A$ as then $$0 \leq \lim\limits_{n \to \infty} a_{n+1} < \lim\limits_{n \to \infty}\left(1 + \frac 1n \right)a_n -1,$$ which implies $A < A-1$

And $a_n$ cannot be bounded by $d$ as we would have, for $n > 2d$, $a_{n+1} < a_n\left(1 + \frac 1n\right) - 1 < d\left(1 + \frac {1}{2d}\right) - 1 = d - \frac 12$. Continuing in this way there would be an $a_n < 0$.

I was thinking of using the fact that the superior limit is the largest partial limit.
What should I do next or should I use a different method?


Let us start from your conclusion $s_{n}<r_n$ for $n\geq M$. This can be written in the following equivalent form $$ \forall\, n\geq M,\quad\frac{1}{n+1}<\frac{a_n}{n}-\frac{a_{n+1}}{n+1} $$ This implies, by adding these inequalities $$ \forall\, m\geq M,\quad\sum_{k=M+1}^{m-1}\frac{1}{k+1}<\frac{a_{M+1}}{M+1}-\frac{a_{m}}{m} <\frac{a_{M+1}}{M+1} $$ and letting $m$ tend to $+\infty$ we get a contradiction. Thus the desired limit superior is larger or equal to $e$.


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