$\limsup\left(\frac{a_1+a_{n+1}}{a_n}\right)^n\ge c$ Let $a_n>0,n\in\mathbb{N}$ be a sequence of positive real numbers. There exists a positive real number $c$ such that $\limsup\left(\frac{a_1+a_{n+1}}{a_n}\right)^n\ge c$ as $n\to\infty$ for all $\{a_n\}$. Find with proof the maximum possible value of $c$.
 A: I will summarize Polya's solution, i.e., a proof that 
$$\limsup_{n \rightarrow \infty} \left ( \frac{a_1+a_{n+1}}{a_n} \right ) \ge e$$
Assume the opposite, i.e.,
$$\limsup_{n \rightarrow \infty} \left ( \frac{a_1+a_{n+1}}{a_n} \right ) < \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^n = e$$
Then 
$$\limsup_{n \rightarrow \infty} \left [ \frac{n(a_1+a_{n+1})}{(n+1) a_n} \right ]^n < 1$$
which implies that $\exists N \in \mathbb{N} : \forall \, n > N$
$$ \frac{n(a_1+a_{n+1})}{(n+1) a_n} < 1 \implies \frac{a_1 + a_{n+1}}{n+1} < \frac{a_n}{n}$$
or
$$\frac{a_{n+1}}{n+1} - \frac{a_n}{n} < -\frac{a_1}{n+1}$$
We may sum terms like this from, say, $n=N$ to some $K>N$ and get
$$\frac{a_{K+1}}{K+1} - \frac{a_N}{N} < -a_1 \sum_{k=N}^{K} \frac{1}{k+1}$$
Note that the sum on the right goes to $-\infty$ as $K \rightarrow \infty$.  Thus,
$$\lim_{K \rightarrow \infty} \frac{a_{K+1}}{K+1} = -\infty$$
which is a contradiction of the fact that $a_K >0$.  Thus, the conjecture is proven.
A: Let me complete the answer of Ron Gordon:
Ron Gordon proved that: For any sequence $(a_n)$ with positive terms:
$$\limsup_{n\to\infty} \Big(\frac{a_1+a_{n+1}}{a_n}\Big)^n\geq c,$$
with $c=e$.
We were also asked for the maximum value $c_{max}$ of such a constant $c$, thus it remains to prove that $e$ is the largest constant $c$ for which this inequality holds for any sequence $(a_n)$ with positive terms.
Indeed by choosing a sequence $(a_n)$ such that
$$a_n=\alpha+\beta (n-1),\quad\alpha\,,\beta>0$$
we get
$$\begin{align}
\big(\frac{a_1+a_{n+1}}{a_n}\big)^n&=\exp\big(n\ln(\frac{\alpha+\alpha+\beta n}{\alpha +\beta (n-1)})\big)\\
&=\exp\big(n\ln(1+\frac{\alpha+\beta}{\alpha+\beta (n-1)})\big)\\
&\leq\exp\big(n\frac{\alpha+\beta}{\alpha+\beta (n-1)}\big)\\
&\leq\exp(\frac{\alpha+\beta}{(\alpha-\beta)/n+\beta})\\
\end{align}$$
and thus $$\limsup_{n\to\infty} \Big(\frac{a_1+a_{n+1}}{a_n}\Big)^n \leq\exp(1+\alpha/\beta)\,,$$ 
and hence, for all $\alpha,\beta>0$,
$$c_{max}\leq\exp(1+\alpha /\beta)$$
which brings the inequality
$$c_{max}\leq e$$
since $\alpha/\beta$ can be chosen as small as we want.
