Upper bound for limsup

Question

Question - Let $\{a_{n}\}$ be a sequence of positive real numbers. Then $\limsup(\frac{1+a_{n}}{a_{n}})^{n}\geq e$.

I don't know how to prove it. I can only think that I need to use that $\lim (\frac{n+1}{n})^{n}= e$.

Answer 1

The statement is false, there is a counterexample:

Choose some number $c$ with $1<c<e$. and define $a_n := \frac{1}{\sqrt[n]{c} - 1}$.

Then $$\left(\frac{1+a_n}{a_n} \right)^n = c < e$$

for all $n$.