Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$ 
Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$.

I have to prove this inequality using induction, I proved it for $n=1$ and now I have to prove it for $n+1$ assuming $n$ as hypothesis, but this seems impossible to me because the difference between the sum of $n$ and the sum of $n+1$ is a positive value. Adding a positive value on both sides of the inequality, I don't know how to prove that is always less than or equal to $\frac{3}{4}$. 

Let
$$S_n=\sum_{i=1}^{n}\frac{1}{n+i}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}=H_{2n}-H_n.\tag{1}$$
We have
$$ S_{n+1}-S_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)} \tag{2}$$
hence $\{S_n\}_{n\geq 1}$ is an increasing sequence. Due to $(1)$ and Riemann sums,
$$S_n \leq \lim_{n\to +\infty}S_n = \int_{0}^{1}\frac{dx}{1+x} = \log(2). \tag{3}$$
Over the interval $(0,1)$, the function $f(x)=x(1-x)$ is positive and $\leq \frac{1}{4}$, hence
$$ \frac{1}{16}\geq \int_{0}^{1}\frac{x^2(1-x)^2}{1+x}\,dx = -\frac{11}{4}+4\log 2\tag{4} $$
and $$\log(2)\leq \frac{45}{64}<\frac{3}{4}.\tag{5}$$
