Limit of $a_{n+1}=a_{n}\cdot \left( 1+\frac{1}{n}\right)$

Question

Let $a_{n}$ be a recursive sequence such as: $\begin{cases}a_{1}=1\\ a_{n+1}=a_{n}\cdot \left( 1+\dfrac{1}{n}\right) \end{cases}$

I need to show that $\lim _{n\rightarrow \infty }a_{n}=\infty$.

I showed that $a_{n}$ is increasing using induction, so we have two options:

1. $\lim _{n\rightarrow \infty }a_{n}=L$
2. $\lim _{n\rightarrow \infty }a_{n}=\infty$

I have tried assuming that $\lim _{n\rightarrow \infty }a_{n}=L$ but I couldn't get to a contradiction.

Would appreciate some help!

Answer 1

If $a_n =n, a_{n+1}=({n+1}/n) n=n+1$

since $a_1=1$ we can conclude that by induction $a_n=n$

Answer 2

It is clear that $ \frac{a_{n+1}}{n+1}=\frac{a_{n}}{n}$ for all $n$. Therefore $\frac{a_{n}}{n}=\frac{a_{1}}{1}=1$. Thus $a_{n}=n$ and we are done!