Evaluate $\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right)$

Question

I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$

I have computed \begin{align*} a_{n+1}-a_n & = \frac{1}{\sqrt{(2n+1)(2n+2)}}-\frac{1}{\sqrt{n(n+1)}}\\ & = \frac{1}{\sqrt{n+1}}\left( \frac{1}{\sqrt{2(2n+1)}}-\frac{1}{\sqrt{n}} \right). \end{align*} Now I'm stuck.

Answer 1

Hint: $$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}<\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)(2n)}}$$

$$<\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}$$ and use $\lim_{n \rightarrow \infty} (\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n})=\ln 2$

Answer 2

HINT

The hint provided by Chinnapparaj R is the key point, to evaluate the bounding sum we can use that

$$\frac1n+\frac1{n+1}+\ldots+\frac1{2n}=\sum_{k=1}^n \frac{1}{n+k}=\frac1n \sum_{k=1}^n \frac{1}{1+\frac kn}$$

which is a Riemann's sum.