Inequality $\sum\limits_{k=1}^n \frac{1}{n+k} \le \frac{3}{4}$

Solution 1:

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!!!

Solution 2:

$$\sum_{k=1}^{n}\frac{1}{n+k}\leq \sum_{k=1}^{n}\frac{1}{\sqrt{(n+k)(n+k-1)}}\stackrel{\text{CS}}{\leq}\sqrt{n\sum_{k=1}^{n}\left(\frac{1}{n+k-1}-\frac{1}{n+k}\right)}$$ immediately leads to $H_{2n}-H_n = \sum_{k=1}^{n}\frac{1}{n+k}\leq \frac{1}{\sqrt{2}}<\frac{3}{4}$.

Solution 3:

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.