Lower bound for the upper limit of the sequence

Question

Prove that for any sequence $ a_n > 0 $ the limit superior of $\left(\frac{1+a_{n+1}}{a_n}\right)^n$ is greater or equal to $ e $.
It looks similar to the definition of $ e $, but I can't figure out where to start from. I think maybe it can be proved by contradiction, but I don't know how exactly.

Answer 1

Suppose $$\frac{1+a_{n+1}}{a_n}\ge1+\frac1n\tag{1}$$ were true only for finitely many values of $n$: that means $$\frac{1+a_{n+1}}{a_n}<1+\frac1n\tag{2}$$ for $n\ge n_0$. But (2) is equivalent to $$ \frac{a_{n+1}}{n+1}<\frac{a_n}n-\frac1{n+1},$$ so by summing that up, we'd have $$\frac{a_{n+1}}{n+1}<\frac{a_{n_0}}{n_0}-\sum^n_{k=n_0}\frac1{k+1}$$ for $n\ge n_0$ Since the harmonic series diverges, we'd have $a_{n+1}<0$ for sufficiently large $n$, contrary to our assumptions. So (1) must be true for infinitely many $n$, and that means $$\limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n\ge\limsup_{n\to\infty}\left(1+\frac1n\right)^n=e.$$