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 $\frac{3}{4}$. 
 A: Preliminary Result
Note that
$$
\begin{align}
s_{n+1}-s_n
&=\sum_{k=1}^{n+1}\frac1{n+k+1}-\sum_{k=1}^n\frac1{n+k}\\
&=\sum_{k=2}^{n+2}\frac1{n+k}-\sum_{k=1}^n\frac1{n+k}\\
&=\frac1{2n+2}+\frac1{2n+1}-\frac1{n+1}\\
&=\frac1{2n+1}-\frac1{2n+2}\tag{1}
\end{align}
$$
Therefore, since $\frac1{2k}-\frac1{2k+1}\ge\frac12\left(\frac1{2k}-\frac1{2k+2}\right)$
$$
\begin{align}
s_n
&=\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2k}\right)\\
&=1-\sum_{k=1}^{n-1}\left(\frac1{2k}-\frac1{2k+1}\right)-\frac1{2n}\\
&\le1-\frac12\sum_{k=1}^{n-1}\left(\frac1{2k}-\frac1{2k+2}\right)-\frac1{2n}\\[3pt]
&=\frac34-\frac1{4n}\tag{2}
\end{align}
$$

Inductive Proof
We will prove inductively that
$$
\sum_{k=1}^n\frac1{n+k}\le\frac34-\frac1{4n}\tag{3}
$$
Note that $(3)$ holds, with equality, for $n=1$.
Suppose that
$$
\sum_{k=1}^n\frac1{n+k}\le\frac34-\frac1{4n}\tag{4}
$$
Note that $\frac1{4n+4}\le\frac1{4n}-\frac1{2n+1}+\frac1{2n+2}\iff\frac1{2n+1}\le\frac12\left(\frac1{2n}+\frac1{2n+2}\right)$, which is true because $\frac1x$ is convex. Therefore,
$$
\begin{align}
\sum_{k=1}^{n+1}\frac1{n+k+1}
&=\sum_{k=2}^{n+2}\frac1{n+k}\\
&=\color{#00A000}{\sum_{k=1}^n\frac1{n+k}}\color{#C00000}{-\frac1{n+1}+\frac1{2n+1}+\frac1{2n+2}}\\
&\le\color{#00A000}{\frac34-\frac1{4n}}\color{#C00000}{+\frac1{2n+1}-\frac1{2n+2}}\\[6pt]
&\le\frac34-\frac1{4n+4}\tag{5}
\end{align}
$$
which completes the induction.
A: Define $\displaystyle{S_n=\sum_{i=1}^n \dfrac{1}{n+i}}$. Then we have for every $n$ that:
\begin{align*}
S_{n+1}-S_n&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i}-\sum_{i=1}^n \dfrac{1}{n+i}\\
&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\sum_{i=2}^n \dfrac{1}{n+i}\right)-\dfrac{1}{n+1}\\
&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\sum_{j=1}^{n-1} \dfrac{1}{n+1+j}\right)-\dfrac{1}{n+1}\\
&=\dfrac{1}{n+1+(n+1)}+\dfrac{1}{n+1+n}-\dfrac{1}{n+1}\\
&=\dfrac{1}{2n+1}-\dfrac{1}{2(n+1)}=\dfrac{1}{2(n+1)(2n+1)}
\end{align*}
In particular, the sequence $S_n$ can also be defined by recurrence with the formula $$\begin{cases}S_1=\dfrac{1}{2}\\ S_{n+1}=S_n+\dfrac{1}{2(n+1)(2n+1)}\end{cases}$$
and so, $S_n$ is simply defined by the formula $$S_n=\sum_{i=1}^n \dfrac{1}{2n(2n-1)}.$$
Finally, showing that $S_n\leq \frac{3}{4}$ for all $n$ is the same as proving that the series $\displaystyle{\sum_{n=1}^\infty \dfrac{1}{2n(2n-1)}}$ converges to a number less than or equal to $\dfrac{3}{4}$.
If we focus on the infinite series, notice that $\dfrac{1}{2n(2n-1)}=\dfrac{1}{2n-1}-\dfrac{1}{2n}$, and so we have:
$$\sum_{n=1}^\infty \dfrac{1}{2n(2n-1)}=\sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n} \right)= \sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$$
I know the last series converges (by the test of alternating series), but I am just not sure about the value. 
EDIT: AS lhf points out in his comment, the value of the last sum is $\log 2=0.69314\ldots$, which is surely less than $\dfrac{3}{4}$.
A: Assuming $n>1$, 
$$\begin{align}
\sum_{i=1}^n \frac 1{n+i}
=\frac 1n \sum_{i=1}^n \frac 1{1+\frac in}
&\color{red}<\int_0^1 \frac 1{1+x} \;\; dx
= \bigg[\ln (1+x)\bigg]_0^1 
=\ln 2=0.693
\color{red}<\frac 34\\
\Rightarrow \sum_{i=1}^n \frac 1{n+i}&\color{red}<\frac 34\qquad\blacksquare\end{align}$$
A: Note that you do not use induction in the above.
Another way:
Define $$S_n := \sum_{i=1}^n \frac{1}{n+i}$$
Then $$S_{n+1} = \sum_{i=1}^{n+1} \frac{1}{n+1+i} = \sum_{i=1}^{n+1} \frac{1}{n+(i+1)} = \sum_{i=2}^{n+2} \frac{1}{n+i}.$$
Hence $$S_{n+1} = S_n - \frac{1}{n+1} + \frac{1}{n+n+1} +\frac{1}{n+n+2}.$$
Now proceed...
A: 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 $$\log(2)\leq \frac{45}{64}<\frac{3}{4}.\tag{5}$$
