Value of $\lim_{n\to \infty}(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}...+\frac{1}{2n})$

Question

What is the value of $\lim_{n\to \infty}(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}...+\frac{1}{2n})$?

Please give some hints to proceed .

Answer 1

Hint: Write $\dfrac{1}{n}+\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n}$ as $\displaystyle\sum_{k = 0}^{n}\dfrac{1}{n+k} = \dfrac{1}{n}\sum_{k = 0}^{n}\dfrac{1}{1+\tfrac{k}{n}}$.

This last expression should look like a Riemann sum. Can you figure out what integral it is a Riemann sum for?

Answer 2

Note that $$\int_{n}^{2n+1}{1\over x}dx>{1\over n} +...+ {1\over 2n}>\int_{n-1}^{2n}{1\over x}dx$$.

You can apply sandwich theorem after that to obtain the limit equal to $\ln2$.