Let $a_n$ be a positive sequence. Prove that $$\limsup_{n\to \infty} \left(\frac{a_1+a_{n+1}}{a_n}\right)^n\geqslant e.$$
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asked Sep 28, 2013 at 13:20
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$\begingroup$ This problem is also present in the book "Problems of mathematical olympiads for university students" (Russian) by Sadovnichiy and Podkolzin. $\endgroup$– njguliyevCommented Sep 28, 2013 at 13:49
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$\begingroup$ A thought: Suppose not. Choose $l<e$ such that $l>\limsup_{n\to\infty}((a_1+a_{n+1})/a_n)^n$, and $N$ such that $(a_1+a_{n+1})/a_n<l^{1/n}$ for all $n\ge N$. I believe that $a_n$ can't be always positive, and the contradiction is on the asymptotic behavior of large $a_n$. $\endgroup$– Yai0PhahCommented Sep 28, 2013 at 14:39
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$\begingroup$ @FrankScience: why $e$ and not $5.7$, then? $\endgroup$– Martin ArgeramiCommented Sep 28, 2013 at 15:35
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2 Answers
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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.
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$\begingroup$ The inequality $e^{1/n}>n/(n-1)$ is not true, unless my calculator and Wolfram Alpha are wrong. Besides, it seems to me that the inequality you use to get your contradiction does not preclude the equality in the original statement, so you cannot expect to get a contradiction. $\endgroup$ Commented Sep 29, 2013 at 16:06
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$\begingroup$ I didn't understand the second part of your comment. $\endgroup$ Commented Sep 29, 2013 at 16:17
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$\begingroup$ I meant that you could have a sequence satisfying $\left((1+a_{n+1})/a_n\right)^n<e$, and such that in the limit you have equality. $\endgroup$ Commented Sep 29, 2013 at 16:49
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$\begingroup$ OK, but I don't see how that contradicts to my solution. My solution actually shows that the case you are talking about is not possible for positive $\{a_n\}$. $\endgroup$ Commented Sep 29, 2013 at 17:13
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$\begingroup$ Nice solution! (+1). Instead of your inequality, You can use $\exp\left(-\sum_{j=N}^{N+k}\frac{1}{j}\right) = \exp\left(-\ln(N+k) + O(1)\right)$ when $k\to+\infty$. $\endgroup$– SiméonCommented Sep 30, 2013 at 8:31
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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 $K \geq 1$.
