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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 $c$.

user60145
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  • I think this problem is in Polya, Mathematics and Plausible Reasoning. However, you may wish to specify where that limit is being taken. – Ron Gordon Jan 29 '13 at 21:35

2 Answers2

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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.

Ron Gordon
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  • Is the step from $\limsup_{n \rightarrow \infty} \left ( \frac{a_1+a_{n+1}}{a_n} \right )^n < \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^n = e$ to $\limsup_{n \rightarrow \infty} \left [ \frac{n(a_1+a_{n+1})}{(n+1) a_n} \right ]^n < 1$ obvious? – Sebastien B Jan 30 '13 at 14:07
  • I think it is self-evident. Can you think of a reason why it is not so? – Ron Gordon Jan 30 '13 at 14:09
  • Well sometime I shall think before to write. It is indeed clear... – Sebastien B Jan 30 '13 at 14:13
  • Technically it should be $\lim_{n\to\infty}\sup_{m\ge n}\left(\frac{a_1+a_{m+1}}{a_m}\right)^m\left(\frac{n}{n+1}\right)^n$, right? Why is this necessarily equal to $\lim_{n\to\infty}\sup_{m\ge n}\left(\frac{(a_1+a_{m+1}m)}{a_m(m+1)}\right)^m$? – Christmas Bunny Feb 07 '13 at 07:04
  • @ChristmasBunny It is a property of the $\limsup$: If $(u_n)$ and $(v_n)$ are two real sequences and $(v_n)$ converges to $v_\infty > 0$, then $\limsup_{n\to\infty} \big{u_n v_n\big} = v_\infty \limsup_{n\to\infty}u_n$. – Sebastien B Dec 30 '14 at 12:21
  • @RonGordon I think the proof that $e$ is the maximum value of the constant $c$ in the inequality is missing. I did that below. – Sebastien B Dec 30 '14 at 13:16
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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.

Sebastien B
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