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

 A: 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.
A: 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 $m.$
