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. 
 A: 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!!!
A: $$\sum_{k=1}^{n}\frac{1}{n+k}\leq \sum_{k=1}^{n}\frac{1}{\sqrt{(n+k)(n+k-1)}}\stackrel{\text{CS}}{\leq}\sqrt{n\sum_{k=1}^{n}\left(\frac{1}{n+k-1}-\frac{1}{n+k}\right)}$$
immediately leads to $H_{2n}-H_n = \sum_{k=1}^{n}\frac{1}{n+k}\leq \frac{1}{\sqrt{2}}<\frac{3}{4}$.
A: All you really need to do is change the lower limit in the integral from $1$ to $0$, which gives the simpler result
$$\sum_{k=1}^n{1\over n+k}\le\int_0^n{dx\over n+x}=\ln(2n)-\ln n=\ln2$$
One way to see that $1$ is the wrong lower limit to use is that, in general, the number of terms in the sum should equal the difference of the upper and lower limits in the integral.  That's because the integral comparison test usually compares each term with the area beneath a curve over a unit segment.  
A: $$\frac{1}{n+k}+\frac1{2n+1-k}=\frac{3n+1}{(n+k)(2n+1-k)}$$
Then show: $$(n+k)(2n+1-k)\geq (n+1)(2n)$$ for $k=1,2,\dots,n.$
This gives:
$$\frac{1}{n+k}+\frac1{2n+1-k}\leq\frac{3n+1}{2n(n+1)}<\frac{3}{2n}$$
If $S$ is your sum, this means that:
$$2S<\frac32,$$ Or $S<\frac34.$

You can use AM/GM to get that:
$$(n+k)(2n+1-k)\leq \left(\frac{3n+1}{2}\right)^2,$$
which gives the lower bound:
$$S\geq\frac{2}{3+\frac{1}n}.$$
