What is the limit of the sum $\frac 1 n+\frac 1 {n+1}+ \cdots+\frac 1 {2n}$?

Question

I am given the following sum $s_n=\displaystyle\sum_{k=1}^{n} \frac{1}{n+k}$.

But I do not know how to obtain its limit as $n\to\infty$.

Answer 1

$$ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}=\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots+\frac{1}{1+\frac{n}{n}}\right)\\ =\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}} \to \int_0^1 \frac{dx}{1+x}=\log 2. $$ Since, for every $f:[a,b]\to\mathbb R$ continuous $$ \frac{b-a}{n}\sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right)\to\int_a^b f(x)\,dx. $$

Answer 2

The harmonic numbers are defined as

$$H_n = \sum_{k=1}^n \frac{1}{k}=\int_0^1 \frac{1 - x^n}{1 - x}\,dx$$

Then your sum is

$$S=H_{2n}-H_{n-1}$$

This is also called a harmonic progression.

Answer 3

If what you're after is the limit of the sequenc$$\left(\frac1n+\frac1{n+1}+\cdots+\frac1{2n}\right)_{n\in\mathbb N},$$then the limit is $\log 2$. You can do it using Riemann sums and noticing that what you are computing is$$\int_0^1\frac{\mathrm dx}{x+1}=\log2.$$