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.

$f(x) = 1/(n+x)$ 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
$$
\sum^{n}_{k=1} \frac{1}{n+k} \le \sum^{n}_{k=1} 1/n + k (1/(2n)- 1/n)/n = \\1 + \frac12 n (n+1) (1/(2n)- 1/n)/n =  1 - \frac14 \frac{n+1}{n} = \frac{3}{4} - \frac{1}{4 n} \le \frac{3}{4}
$$
