# An inequality from Littlewood's Miscellany

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

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

• Littlewood credits Polya with this this result. Jun 11, 2013 at 15:39

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 $m.$

• But maybe that sequence hasn't limit.
– Mher
Jun 11, 2013 at 15:52
• we do not need s limit, just the upper limit. Jun 11, 2013 at 15:53
• What happened to $c$? Jun 11, 2013 at 15:54
• @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 $n$. Jun 11, 2013 at 15:57
• 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. Jun 11, 2013 at 16:15

This is Leshik's proof, but spelled out a bit more.

Lemma 1: If $$0 then for sufficiently large $$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 $$n$$, $$c_n.
Proof: This is practically the definition of $$\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.

• 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, 2021 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, 2021 at 17:41
• @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.” Apr 26, 2021 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, 2021 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.. May 28, 2021 at 2:05