Partial reciprocal sum How can I show that $$\sum_{k=1}^{n}\frac{1}{n+k}\leq\frac{3}{4}$$ for every integer $n \geq 1$?
I tried induction, estimates with logarithms and trying to bound the sum focusing on the larger terms or things like $\frac{1}{n+1}+\frac{1}{n+2}\leq\frac{2}{n+1}$ but nothing seems to work. Do you have any suggestion? Thanks

Here is an elementary proof. Let $$ u_n = \sum_{k=1}^n \frac{1}{n+k}$$
For all $n \geq 1$, we have
$$u_{n+1}-u_n = \sum_{k=1}^{n+1} \frac{1}{n+1+k} - \sum_{k=1}^n \frac{1}{n+k}$$ $$= \sum_{k=2}^{n+2} \frac{1}{n+k} - \sum_{k=1}^n \frac{1}{n+k} = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1}$$ $$ = \frac{1}{2n+1}-\frac{1}{2n+2} = \frac{1}{(2n+1)(2n+2)} > 0.$$
So the sequence $(u_n)_{n \geq 1}$ is strictly increasing.
Moreover, you have
$$u_n = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \frac{k}{n}} \longrightarrow \int_0^1 \frac{1}{1+x} \mathrm{dx} = \ln(2)$$
So the sequence is increasing and converges to $\ln(2)$, so you have $u_n \leq \ln(2)$ for all $n \geq 1$. You deduce
$$u_n \leq \frac{3}{4}$$
