Prove that $ \frac{1}{1+n^2} < \ln(1+ \frac{1}{n} ) < \frac {1}{\sqrt{n} }$ For $n >0$ ,
Prove that $$ \frac{1}{1+n^2} < \ln(1+ \frac{1}{n} ) < \frac {1}{\sqrt{n}}$$
I really have no clue. I tried by working on  $ n^2 + 1 > n  > \sqrt{n} $ but it gives nothing.
Any idea?

Note that 
$$\frac{1}{1+n^2} < \ln\left(1+ \frac{1}{n} \right)< \frac {1}{\sqrt{n}}\iff\frac{n^2}{1+n^2} < n\ln\left(1+ \frac{1}{n} \right)^n < \frac {n^2}{\sqrt{n}}=n\sqrt n$$
which is true, indeed for $n=1$ the given inequality is true and for $n>1$
$$\frac{n^2}{1+n^2} <1< n\ln\left(1+ \frac{1}{n} \right)^n\leq n\log e=n< n\sqrt n$$
