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?
 A: 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 $n>1$
$$\frac{n^2}{1+n^2} <1< n\ln\left(1+ \frac{1}{n} \right)^n\leq n\log e=n< n\sqrt n$$
A: This inequality is correct only for $n >1$ and not for $n >0$ as asked by the person.
For $n >1$, we know that $$\tag1 \frac{1}{n}<\frac{1}{\sqrt n}$$ and $$\tag2 \frac{1}{(1+n^2)} < \frac{1}{(1+n)}$$
Also, $\tag3\frac1{n+1}<\ln\left(1+\tfrac1n\right)<\frac1n, \forall n>0$
So the overall inequality $$\tag4\frac{1}{(1+n^2)} < \frac{1}{(1+n)}<\ln{\left(1+\frac1n\right)}<\frac{1}{n}<\frac{1}{\sqrt n}$$ will be valid only for $n>1$ and not for $n>0$.
A: I found that with 
$$ ln(1+x) < x $$  $\forall x\in ]0,+\infty[$
we can easily prove the RLH side of inequality, because $sqrt(x) < x$
still have the problem of the first one.
oh I found something else :
we have also 
$$ \frac{x}{(1+x)} < ln(1+x) $$  $\forall x\in ]0,+\infty[$
thus here we have $$ ln(1+1/n) >\frac{1}{(1+n)} > \frac{1}{(1+n^2)} $$
I think I have answered my question. thank you :) 
