Evaluate limits by interpreting sums as integral sums Problem: Evaluate the following limits by interpreting given sums as integral sums for certain functions and by using the Fundamental Theorem of Calculus. (a) Find $\lim{S_{n}}$ as n goes to infinity where $$S_{n} = \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} $$
I do not know what this means. My "gut" guess is that I shall evaluate it is a Riemann-sum and the function to be integrated is $$\int_{0}^{x}\frac{1}{1+x} $$
 A: We can rewrite our sum $S_n$ as 
$$S_n=\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+\frac{1}{1+\frac{3}{n}}+\cdots +\frac{1}{1+\frac{n}{n}}\right).$$
Note that $\dfrac{1}{1+\frac{k}{n}}$ is the value of the function $f(x)=\frac{1}{1+x}$ at the point $x=\frac{k}{n}$.
So we have evaluated the function $f(x)$ at equally-spaced points $\frac{1}{n}$, $\frac{2}{n}$, and so on up to $\frac{n}{n}$, multiplied the value by the length $\frac{1}{n}$ of the interval from $\frac{k-1}{n}$ to $\frac{k}{n}$, and added up, $k=1$ to $n$.
So our sum $S_n$ is, for large $n$, a good approximation to the area under $y=\frac{1}{1+x}$, from $x=0$ to $x=1$. As $n\to\infty$, the sum approaches the area, which is $\int_0^1 \frac{1}{1+x}\,dx$. This area is $\ln 2$. 
Remark: The Fundamental Theorem of Calculus comes into consideration because the required area can be found by finding an antiderivative $F(x)=\ln(1+x)$ of $\frac{1}{1+x}$, and calculating $F(1)-F(0)$.
We could write the answer in a different way. For example, let $g(x)=\frac{1}{x}$. We can think of our sum $S_n$ as obtained by evaluating $g(x)$ at $x=1+\frac{1}{n}$, $1+\frac{2}{n}$, and so on up to $1+\frac{n}{n}$, multiplying the result by $\frac{1}{n}$, and adding up. From that point of view, our sum is a Riemann sum for $\int_1^2\frac{1}{x}\,dx$. So $\lim_{n\to\infty}S_n=\int_1^2 \frac{1}{x}\,dx$.  
A: Could it be
$$
\int_0^x\frac1{a+x}\mathrm{d}x/a<=x
$$
? check it out ;)
