Inequality $\sum\limits_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$ I recently came across the following exercise: 

Prove that $$\sum_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$$
  for every natural number $n \ge 1$. 

I immediately tried by induction, by I did not succeed as - after some trivial algebraic manipulations - I arrive at $\sum_{k=0}^{n+1} \frac{1}{n+k} \le \frac{3}{4} + \text{something non-negative}$. 
Thus the solution I have found goes via comparison with the integral of $1/x$: more precisely, 
$$
\sum_{k=1}^n \frac{1}{n+k} \le \int_1^n \frac{dx}{n+x} =\ln(2n) - \ln(n+1) = \ln(2) + \ln\left(\frac{n}{n+1}\right)
$$
and the conclusion follows easily as $\ln 2 \le 3/4$ and the last term is non-positive. 
I would like to ask first if my solution is correct; secondly, do you have other solutions? I believe it should be very easy to prove. 

By C-S
$$\sum_{k=1}^n\frac{1}{n+k}=1-\sum_{k=1}^n\left(\frac{1}{n}-\frac{1}{n+k}\right)=1-\frac{1}{n}\sum_{k=1}^n\frac{k}{n+k}=$$
$$=1-\frac{1}{n}\sum_{k=1}^n\frac{k^2}{nk+k^2}\leq1-\frac{\left(\sum\limits_{k=1}^nk\right)^2}{n\sum\limits_{k=1}^n(nk+k^2)}=1-\frac{\frac{n(n+1)^2}{4}}{n\cdot\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=$$
$$=\frac{7n-1}{2(5n+1)}<0.7\leq\frac{3}{4}.$$
Done!
I think it's interesting that $\ln2=0.6931...$.
C-S forever!!!
