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Let $\{a_n\}$ be sequence of positive terms. Prove that $\displaystyle \lim_{n\to\infty}\sup\left(\frac{a_1+a_{n+1}}{a_n}\right)^n\ge e$

I'm tring to reduce the LHS to some form of the type $\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n$ and also tried using the fact that $\lim \sup a_n\ge \lim a_n$ but couldn't get much.

Pierre-Yves Gaillard
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Mathronaut
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    One problem to think about is that we don't know if $\lim a_{n}$ even exists. The problem states ${ a_{n} }$ is a sequence of positive numbers, but it otherwise gives no indication of the behavior of the sequence. So you have to be careful when saying $\lim{\sup{ a_{n}}} \geq \lim a_{n}$, because the inequality might not make sense if $\lim a_{n}$ doesn't exist. – layman Aug 09 '14 at 17:24
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    @user46944 Indeed, and if $\lim a_n$ exists, the $\limsup a_n =\lim a_n$. – Thomas Andrews Aug 09 '14 at 17:28

1 Answers1

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Consider the set $\mathfrak N$ of integers $n$ such that $\dfrac{a_1+a_{n+1}}{a_n}\lt\dfrac{1+n}n$. For every $n$ in $\mathfrak N$, one has $\dfrac{a_n}n-\dfrac{a_{n+1}}{n+1}\gt\dfrac{a_1}{n+1}$. Assume that $\mathfrak N$ contains every integer after some $n_0$, then, for every $k\geqslant n_0$, $$\dfrac{a_{n_0}}{n_0}\gt\dfrac{a_{n_0}}{n_0}-\dfrac{a_{k}}{k}\gt a_1\sum\limits_{i=n_0+1}^{k}\dfrac1{i}.$$ The lower bound diverges when $k\to\infty$ hence the hypothesis is absurd, that is, the set of $n$ such that $\dfrac{a_1+a_{n+1}}{a_n}\geqslant\dfrac{1+n}n$ is infinite. In particular, $\left(\dfrac{a_1+a_{n+1}}{a_n}\right)^n\geqslant\left(\dfrac{1+n}n\right)^n$ infinitely often. The limit of the RHS when $n\to\infty$ is $\mathrm e$, hence this proves the result.

Likewise, for every positive $c$, $$\limsup_{n\to\infty}\left(\dfrac{c+a_{n+1}}{a_n}\right)^n\geqslant\mathrm e.$$

Did
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