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Why is the following inequality from Chapter 1 of Littlewood's A Mathematician's Miscellany true?

Suppose $a_n>0$ for all $n$. Then $$\limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n\geq e.$$

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    $\begingroup$ Littlewood credits Polya with this this result. $\endgroup$ Jun 11, 2013 at 15:39

2 Answers 2

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Suppose that the upper limit is $c<e.$ Then for sufficiently large $n$ and $b<e$ we have $$\frac{1+a_{n+1}}{a_n}\le b^{1/n}\le {\frac{n+1}{n}}.$$ Introducing $b_n=\frac{a_n}{n},$ we can rewrite the last inequality as $$b_{n+1}+\frac{1}{n+1}\le b_n.$$ Iterating last inequality we arrive at $$b_{n+m}\le b_n-\sum_{k=1}^m\frac{1}{n+k}.$$ The only thing left is to note that harmonic series diverges and the right hand side can be made negative for sufficiently large $m.$

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  • $\begingroup$ But maybe that sequence hasn't limit. $\endgroup$
    – Mher
    Jun 11, 2013 at 15:52
  • $\begingroup$ we do not need s limit, just the upper limit. $\endgroup$
    – leshik
    Jun 11, 2013 at 15:53
  • $\begingroup$ What happened to $c$? $\endgroup$ Jun 11, 2013 at 15:54
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    $\begingroup$ @MherSafaryan See my pevious ocmment - OR note that the negation of $\limsup x_n\ge e$ is: There exists $b<e$ such that $x_n<b$ for almost all $n$. $\endgroup$ Jun 11, 2013 at 15:57
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    $\begingroup$ Leaving something to the reader and skipping past the definition of a variable are very different things, especially when the reader has no idea if you are right. $\endgroup$ Jun 11, 2013 at 16:15
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This is Leshik's proof, but spelled out a bit more.

Lemma 1: If $0<x<e$ then for sufficiently large $n$, $x^{1/n}<1+1/n$.
Proof: I'll leave this for you for now.

Lemma 2: Given any sequence $c_n$, if $\limsup c_n < x$ then for sufficiently large $n$, $c_n<x$.
Proof: This is practically the definition of $\limsup$.

Main proof: Let $c_n = \left(\frac{1+a_{n+1}}{a_n}\right)^n$. We proceed to prove by contradiction. Assume $\limsup c_n < e$. Then let $b$ be some value such that $\limsup c_n < b< e$.

Now, for large enough $n$, $b > c_n=\left(\frac{1+a_{n+1}}{a_n}\right)^n$ by lemma 2, and $b^{1/n}<1+1/n$, by lemma 1, which gets us to the step that Leshik gets:

$$\frac{1+a_{n+1}}{a_n} < b^{1/n} < 1+1/n$$

for sufficiently large $n$. The rest of Leshik's proof is fairly clear - you proceed to show that some $a_i$ must be negative, thus reaching a contradiction, and so our assumption cannot be true, and you are done.

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  • $\begingroup$ Sir thanks a lot for this. It is very helpful indeed +1. I'd been trying to solve it for a long time. I think that Lemma 2 needs one small clarification: it is not true just for sufficiently large $n$ rather for sufficiently large values of $n$. Anyways, I was wondering if it could be proven by taking log and simplifying fractions to summations. I tried with that but didn't get contradiction. $\endgroup$
    – Koro
    Apr 26, 2021 at 14:19
  • $\begingroup$ The difference between "sufficiently large value of $n$" and "sufficiently large values of $n$" is basically "$\exists $N_1$ $ such that statement is true for $n=N_1$" and "$\exists N: n\ge N\implies $ statement is true for all $n\ge N$" respectively. $\endgroup$
    – Koro
    Apr 26, 2021 at 17:41
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    $\begingroup$ @Koro Personally, I take “for sufficiently large $n$“ to always mean that, for some $N$, all $n>N.$ If it implied only one such $n,$ then your term would only imply more than one such $n,$ which would also be misleading. And I didn’t say “value.” $\endgroup$ Apr 26, 2021 at 17:41
  • $\begingroup$ In that case, I don't have any issue. It seems that both terms are misleading "for sufficiently large values of $n$" or "sufficiently large value of $n$" however if the idea behind the usage of these terms is made explicit then there is no issue/ ambiguity at all. $\endgroup$
    – Koro
    Apr 26, 2021 at 17:46
  • $\begingroup$ I don't quite understand why it has to be $\frac{1+a_{n+1}}{a_{n}}$? Can it just be $\frac{1+a_{m}}{a_{n}}$? It looks like $\{a_{n}\}$ is quite arbitrary.. $\endgroup$
    – Shine
    May 28, 2021 at 2:05

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