How we can prove that: $\sum _{k=n}^n f\left(\frac{k}{n}\right)\le n\cdot \log(2)$? $f:\left[0,1\right]\rightarrow R,\:f(x)=\frac{1}{1+x}$ and we have to show that $\sum_{k=n}^n f\left(\frac{k}{n}\right)\le n\cdot\log(2)$. What I know is just that: $n\cdot \log(2)=\int_0^1 \frac{1}{1+x} \, dx$, but don't have idea how can I prove inequality...
$$\:\frac{1}{n} \sum_{k=1}^n\frac{1}{1+\frac{k}{n}}=\int_0^1 \frac{dx}{1+x} = \log 2,$$
form of Riemann, and why
$$\frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}} <\int _0^1\frac{dx}{1+x}=\log 2\text{ ?}$$
 A: Maybe you want to prove:
$$ \sum_{k=\color{red}{1}}^{n}\frac{1}{1+\frac{k}{n}}\leq n\log 2\tag{1}$$
that follows from:
$$ \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\leq\int_{0}^{1}\frac{dx}{1+x}=\log 2 \tag{2}$$
since $f(x)=\frac{1}{1+x}$ is a Riemann-integrable, decreasing function over $[0,1]$. 
If you do not like the integral approach, you may also notice that $(1)$ is equivalent to:
$$ \sum_{k=n+1}^{2n}\frac{1}{k}\leq \log 2 \tag{3}$$
that can be proved through:
$$ \sum_{k=n+1}^{2n}\frac{1}{k}\leq\sum_{k=n+1}^{2n}\operatorname{arctanh}\frac{1}{k}=\frac{1}{2}\log\prod_{k=n+1}^{2n}\frac{1+\frac{1}{k}}{1-\frac{1}{k}}=\frac{1}{2}\log\frac{4n+2}{n+1}\leq \log 2.\tag{4}$$
A: We have $$ \sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}=\sum_{k=1}^{n}\frac{n}{n+k}=n\sum_{k=n+1}^{2n}\frac{1}{k}=n\left(H_{2n}-H_{n}\right)
 $$ where $H_{n}
 $ is the $n
 $-th harmonic number and using the bounds $$\log\left(n\right)+\gamma+\frac{1}{2\left(n+1\right)}<H_{n}<\log\left(n\right)+\gamma+\frac{1}{2n}$$ we get $$\leq n\left(\log\left(2n\right)+\gamma+\frac{1}{4n}-\log\left(n\right)-\gamma-\frac{1}{2\left(n+1\right)}\right)\leq n\log\left(2\right).$$
