Let $a_n>0$. Prove that $$\varlimsup_{n\to\infty}n\left(\frac{1+a_{n+1}}{a_n}-1\right)\geq 1.$$
I argue by contradiction. If it is not ture, then $$\exists\ N,\ \forall\ n\geq N, n\left(\frac{1+a_{n+1}}{a_n}-1\right)<1\Rightarrow 1+a_{n+1}-a_n<\frac{a_n}{n}.$$ What contradiction shall I get then....
Suppose to the contrary that for $n\ge N$, $$ n\left(\frac{1+a_{n+1}}{a_n}-1\right)\lt1 $$ Then, for $n\ge N$, $$ \frac{a_{n+1}}{n+1}\lt\frac{a_n}{n}-\frac1{n+1} $$ It can be shown by induction that $$ \frac{a_{n+k}}{n+k}\lt\frac{a_n}n-\sum_{j=1}^k\frac1{n+j} $$ Since the harmonic series diverges, at some point $\dfrac{a_{n+k}}{n+k}<0$, contrary to the hypothesis that $a_n\gt0$.