Show that $ x_n = \left(1 + \frac{r}{n} \right)^n $ has an upper bound

Question

I asked this question but maybe my doubt was not enough clear. So I will ask something more specific: Show the sequence $x_n = \left(1 + \frac{r}{n}\right)^n$ for $ r \in \mathbb{Q}, r>0$ has an upper bound.

I tried to show it as we do for $a_n = \left(1 + \frac{1}{n} \right)^n $ (using the expansion of $a_n$ but I had no sucess.

I would like some hint!

Thanks!

Answer 1

If you know that $\log(x)\leq x-1$ which is clear from the symmetry with the inverse function $\text e^x\geq x+1$ it follows that $$ 0\leq \log(x_n)=n\ \log\left(1+\frac{r}{n}\right)\leq r $$ showing that $x_n$ is bounded. In fact this shows that $1\leq x_n\leq \text{e}^r$. Also it should be noted here that $\log(x)$ is an increasing function thus preserving inequalities.