Limit superior inequalities proof: $\limsup_{n\to \infty} \left(\frac{a_1+a_{n+1}}{a_n}\right)^n\ge e$ Let $a_n$ be a positive sequence. Prove that 
$$\limsup_{n\to \infty} \left(\frac{a_1+a_{n+1}}{a_n}\right)^n\geqslant e.$$
 A: Since I solved this problem several years ago, I didn't write my solution immediately, so that others could think on this problem. Now I am writing my own solution:
It starts as the solution by Ju'x, i.e. we can safely assume that $a_1 = 1$ and suppose the converse inequality. Then there exists $N \in \mathbb{N}$ such that 
$$\frac{1+a_{n+1}}{a_n} < e^{1/n}, \qquad n \ge N.$$
Hence
$$a_N > \frac{1}{e^{1/N}} + \frac{a_{N+1}}{e^{1/N}} > \frac{1}{e^{1/N}} + \frac{1}{e^{(1/N)+(1/N+1)}} + \frac{a_{N+2}}{e^{(1/N)+(1/N+1)}} > \ldots,$$
i.e.
$$a_N > \frac{1}{e^{1/N}} + \frac{1}{e^{(1/N)+(1/N+1)}} + \ldots + \frac{1}{e^{(1/N)+\ldots+(1/N+k)}}, \qquad k \in \mathbb{N}.$$
Using $e^{1/n} < 1 + \dfrac{1}{n-1} = \dfrac{n}{n-1}$ we get
$$a_N > (N-1)\left( \frac{1}{N} + \frac{1}{N+1} + \ldots + \frac{1}{N+k}\right), \qquad k \in \mathbb{N},$$
which is impossible, since the harmonic series diverges.
A: Without loss of generality we can assume $a_1=1$.
Taking logarithms and seeking for a contradiction, suppose that there exists $0 < \alpha < 1$ such that for all $n$ large enough,
$$
\ln \left(\dfrac{1+a_{n+1}}{a_n}\right) = \ln a_{n+1} - \ln a_n + \ln\left(1+\frac{1}{a_{n+1}}\right)\leq \frac{\alpha}{n}
$$
From this inequality we deduce that $\ln a_{n+1} - \ln a_n  \leq  q/n$.
Summing up, this yields $\ln a_n \leq \alpha\ln n + O(1)$ so $\fbox{$a_n \leq C\,n^\alpha$}$ for some $C > 0$.
Assuming (see below) that we can prove $\lim a_n = +\infty$, we find
$$
\ln a_{n+1} + S_n - T_n \leq \ln a_{n+1} + \sum_{k=1}^{n+1}\ln\left(1+\frac{1}{a_k}\right) \leq \alpha \ln n + O(1)
$$
with
$$
S_n = \sum_{k=1}^n \frac{1}{a_k},\qquad T_n = \sum_{k=1}^n \frac{1}{a_k^2}.
$$
This is absurd because $T_n$ is negligible with respect to $S_n$ and
$$
S_n \geq \frac{1}{C} \sum_{k=1}^n\frac{1}{k^\alpha} \sim \frac{1}{C(1-\alpha)}n^{1-\alpha}.$$

In order to prove that $\lim a_n =+\infty$, start from
$$
\ln a_n \geq -\frac{\alpha}{n} + \ln(1+a_{n+1}) \geq -\frac{\alpha}{n}
$$
This gives $\liminf a_n \geq \lim e^{-\alpha/n} = 1$.
Then write $\liminf a_n \geq \liminf e^{-\alpha/n}(1+a_{n+1}) \geq 2$, and so on... you can show that $\liminf a_n \geq K$ for every integer $K \geq 1$.
