$\limsup\left(\frac{a_1+a_{n+1}}{a_n}\right)^n\ge c$

Question

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 $c$.

Answer 1

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.

Answer 2

Just a partial answer:

The maximum value $c_{max}$ for such a $c$ is smaller than $\exp(1)$.

Indeed by choosing a sequence $(a_n)$ such that $$a_n=\alpha+\beta n,\quad\alpha\,,\beta>0$$ we get $$\begin{align} \big(\frac{a_1+a_{n+1}}{a_n}\big)^n&=\exp\big(n\ln(\frac{2\alpha+\beta(n+1)}{\beta n})\big)\\ &=\exp\big(n\ln(1+\frac{2\alpha+\beta}{\beta n})\big)\\ &\leq\exp\big(n\frac{2\alpha+\beta}{\beta n}\big)\\ &\leq\exp(\frac{2\alpha+\beta}{\beta})\\ \end{align}$$ and hence, for all $\alpha,\beta>0$, $$c_{max}\leq\exp(1+2\alpha/\beta)$$ which brings the inequality $$c_{max}\leq \exp(1)$$ since $2\alpha/\beta$ can be chosen as small as we want.