Show that ${\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$ 
Prove that $\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$ for each positive integer $n$.

My work. I think that i have to use induction, but i can't see how... What i did:
$$f(n)=\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \implies k(n)=f(n+1)-f(n)=\frac{1}{2n+1}-\frac{1}{2n+2}.$$
Now we have 
$$f(n+1)=\displaystyle{\sum^{n}_{k=1}} \left ( \frac{1}{2k+1} \right )- \displaystyle{\sum^{n}_{k=1}} \left ( \frac{1}{2k+2} \right )$$
But now I have no more ideas.
 A: Here there are two different proofs.
1) Note that since $1/(1+x)$ is decreasing and positive for $x\geq 0$ then
$$\sum^{n}_{k=1} \frac{1}{n+k}=\frac{1}{n}\sum^{n}_{k=1} \frac{1}{1+\frac{k}{n}}\leq
\sum^{n}_{k=1}\int_{(k-1)/n}^{k/n} \frac{dx}{1+x}
= \int_0^1 \frac{dx}{1+x}=\ln(2)<\frac{3}{4}.$$
2) For $n\geq 1$, show by induction the stronger inequality (which turns out also in  Andreas' answer)
$$\sum^{n}_{k=1} \frac{1}{n+k} \le \frac{3}{4}-\frac{1}{4n}.$$
In fact, it holds for $n=1$ and for $n\geq 1$,
\begin{align*}
\sum^{n+1}_{k=1} \frac{1}{n+1+k}
&=\sum^{n+2}_{j=2} \frac{1}{n+j}=\sum^{n}_{j=1} \frac{1}{n+j}-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\\
&\leq \frac{3}{4}-\frac{1}{4n}-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\\
&= \frac{3}{4}-\frac{1}{4(n+1)}-\frac{1}{4n(n+1)(2n+1)}
\leq \frac{3}{4}-\frac{1}{4(n+1)}.
\end{align*}
A: $f(x) = 1/(n+x)$ is convex in $x$. So we have that the value for $f(x)$ for any $x \in (0 \; n]$ is less or equal to (line equation between the two interval boundaries) $1/n + x (1/(2n)- 1/n)/n $. Summing the values gives
$$
\sum^{n}_{k=1} \frac{1}{n+k} \le \sum^{n}_{k=1} 1/n + k (1/(2n)- 1/n)/n = \\1 + \frac12 n (n+1) (1/(2n)- 1/n)/n =  1 - \frac14 \frac{n+1}{n} = \frac{3}{4} - \frac{1}{4 n} \le \frac{3}{4}
$$
A: With the partial sums of the harmonic series $H_n$, we have
$$\sum^n_{k=1}\frac1{n+k}=H_{2n}-H_n=\sum^{2n}_{k=1}\frac{(-1)^{k-1}}{k}=\sum^n_{k=1}\left(\frac1{2k-1}-\frac1{2k}\right)=\sum^n_{k=1}\frac1{4k^2-2k}.$$ Since
$$4k^2-2k=4(k^2-k/2)\ge4k(k-1),$$ we can write
$$\sum^n_{k=1}\frac1{4k^2-2k}=\frac12+\sum^n_{k=2}\frac1{4k^2-2k}\le\frac12+\frac14\sum^n_{k=2}\left(\frac1{k-1}-\frac1k\right)=\frac34-\frac1{4n}<\frac34.$$
Interestingly, that's the same result as in two other answers, though the methods are different.
