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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 $$\limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n\geq e.$$

Did
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ryang
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2 Answers2

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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 $m.$

leshik
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  • But maybe that sequence hasn't limit. – Mher Jun 11 '13 at 15:52
  • we do not need s limit, just the upper limit. – leshik Jun 11 '13 at 15:53
  • What happened to $c$? – Thomas Andrews Jun 11 '13 at 15:54
  • well, since $(1+1/n)^n$ converges to $e,$ then for each $b<e$ starting from some $n$ the inequality mentioned above has to be valid. – leshik Jun 11 '13 at 15:55
  • your $c$ is the limit or upper limit..? – Mher Jun 11 '13 at 15:55
  • It should be added that $b=\frac{e+c}2$ for example (or any value strictly between $c$ and $e$. – Hagen von Eitzen Jun 11 '13 at 15:55
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    @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 $n$. – Hagen von Eitzen Jun 11 '13 at 15:57
  • yes, we assume $c\le b<e.$ – leshik Jun 11 '13 at 15:57
  • If $c=b$ then you don't know that inequality. You need $c<b$. @leshik . – Thomas Andrews Jun 11 '13 at 16:00
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    i claim that such $b$ exists, so I do not need to request the strict inequality. – leshik Jun 11 '13 at 16:02
  • But in a proof, we don't "claim," we show. It would be easy to edit this proof to make it clearer and an excellent answer, but you jump around. You don't even assert "there exists a $b$," you just suddenly use the letter $b$ as if the reader knows what you are saying. – Thomas Andrews Jun 11 '13 at 16:06
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    I usually tend to leave some details for the reader. In case the proposer or someone else has questions, I am always willing to clarify things. I can see no point been picky about smth. when you actually understand what is going on, unless something is written in a wrong way. – leshik Jun 11 '13 at 16:13
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    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. – Thomas Andrews Jun 11 '13 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.

Thomas Andrews
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  • 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. – Koro Apr 26 '21 at 14:19
  • 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. – Koro Apr 26 '21 at 17:41
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    @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.” – Thomas Andrews Apr 26 '21 at 17:41
  • 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. – Koro Apr 26 '21 at 17:46
  • 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.. – Shine May 28 '21 at 02:05