# A question on limsup

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....

• i think there are some typo mistakes, or i can't figure out what are $a_n$ and why the 2 limits are written differently ( first is divided by $n$, second $a_n$ – Riccardo May 18 '13 at 13:38
• @RicPed Let $a_n>0$ just means that the sequence we consider is positive. – XLDD May 18 '13 at 13:52
• yeah right, didn't pay attention :( – Riccardo May 18 '13 at 13:53
• But, is it $n$ in the nominator or $a_n$???? – Berci May 18 '13 at 14:51
• @Berci Probably $a_n$, otherwise the statement is not correct (e.g. $a_n := \frac{1}{2}$). – saz May 18 '13 at 14:53

## 1 Answer

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 $a_n\gt0$.