Why is the Riemann sum less than the value of the integral? Why is $ \frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}\leq\int_0^1 \frac{dx}{1+x}=\log 2 $?
Because I think: $$\int _0^1\frac{dx}{1+x}=\frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}}$$
Why is the Riemann sum less than the value of the integral?
 A: Because when $f$ is decreasing, then
$$
\frac 1n f\left(\frac kn\right)\le \int_{\frac{k-1}n}^{\frac kn} f(t)dt
$$
A: Expanding on comments above to give an intuitive idea of this result: write $f(x) = \frac{1}{1+x}$ and set $n = 4$. A rough diagram:

Then, for example, the area of the first (left most) rectangle is 
$$f(1+1/4) \cdot \Delta x = \frac{1}{1+1/4} \cdot \frac{1}{4}$$
The area of all four rectangles is
$$\sum_{k=1}^4 f(1 + k/4) \Delta x = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n}$$
That area is strictly less than $\displaystyle \ \int_0^1 f(x) \ dx$; there is positive area in those 'white gaps' in between the top of the rectangles and the graph of $f$.

It shouldn't be hard to convince yourself intuitively that for all $n$,
$$ \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n} < \int_0^1 f(x) \ dx$$
Here's how I would prove it: 
$f$ is strictly decreasing on $[0,1]$ as $f'(x) = -1/(1+x)^2 < 0$. 
Thus on any interval $[(k-1)/n,k/n]$ for $k =1, 2, \cdots, n$, the minimum value of $f$ on that interval is $f(k/n)$. Hence
$$\int_{(k-1)/n}^{k/n} f(x) \ dx > \int_{(k-1)/n}^{k/n} f(k/n) \ dx = \int_{(k-1)/n}^{k/n} \frac{1}{1+k/n} \ dx = \frac{1}{n} \cdot \frac{1}{1 + k/n}$$
Therefore
$$ \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n} < \sum_{k=1}^n \int_{(k-1)/n}^{k/n} f(x) \ dx = \int_0^1 f(x) \ dx$$
A: The sequence
$$I_n=\frac1n\sum_{k=1}^n\frac1{1+\frac kn}$$
is strictly decreasing and
$$\lim_{n\to\infty}I_n=\int_0^1\frac{dx}{1+x}$$
Hence, the integral is lesser than the Riemann sums.
To show that the sequence is decreasing:
$$I_{n+1}-I_n=\frac1{2n+2}+\frac1{2n+1}-\frac1n<\frac1{2n}+\frac1{2n}-\frac1n=0$$
