How to prove $n < \left(1+\frac{1}{\sqrt{n}}\right)^n$ I want to know how to prove the following inequality.
For $n = 1, 2, 3, \ldots $
$$ n < \left(1+\frac{1}{\sqrt{n}} \right)^n $$
I tried with math induction but I failed. 

We have that
$$n < \left(1+\frac{1}{\sqrt{n}} \right)^n\iff \ln n< n\ln \left(1+\frac{1}{\sqrt{n}} \right)\iff \frac{\ln n}{n}<\ln \left(1+\frac{1}{\sqrt{n}} \right).$$
Now, we use the inequalities (see $(3)$ in http://ajmaa.org/RGMIA/papers/v7n2/pade.pdf)
$$\begin{equation}\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.\tag{1}\end{equation}$$
In our case, we get from $(1),$
$$\ln \left(1+\frac{1}{\sqrt{n}} \right)\ge \frac{2}{\sqrt{n}+2}.$$
Thus we only have to show 
$$\frac{2}{\sqrt{n}+2}>\frac{\ln n}{n}.$$ That is,
$$\ln n<\frac{2n}{\sqrt{n}+2}.$$ Since $\ln n=2\ln \sqrt{n}$ previous inequality is equivalent to
$$\ln \sqrt{n}<\frac{n}{\sqrt{n}+2}.$$ Now, using again $(1)$
$$\ln \sqrt{n}=\ln (1+\sqrt{n}-1)\le \frac{\sqrt{n}-1}{2}\frac{\sqrt{n}+1}{\sqrt{n}}=\frac{n-1}{2\sqrt{n}}.$$ Thus we have to show
$$\frac{n-1}{2\sqrt{n}}<\frac{n}{\sqrt{n}+2}.$$ But
$$\frac{n-1}{2\sqrt{n}}<\frac{n}{\sqrt{n}+2}\iff 2(n-1)<(n+1)\sqrt{n},$$ which holds for any $n\in\mathbb{N}.$
