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

Problem: Evaluate $$ \lim_{n\to\infty}\left(\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+\cdots+\frac{1}{\sqrt{n}\sqrt{n+n}}\right). $$ I have decent familiarity with limits, but I don't see how integrals are related to this problem at all (this problem is in the section on integrals in my textbook). It looks to me like this limit would be $0$ or is that off base?

Solution 1:

$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+\cdots+\frac{1}{\sqrt{n}\sqrt{n+n}}$$ $$=\lim_{n\rightarrow\infty}\sum_{i = 1}^n\frac{1}{\sqrt{n}\sqrt{n+i}} = \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i = 1}^n\frac{1}{\sqrt{1+i/n}} = \int_0^1 \frac{1}{\sqrt{1 + x}} = 2(\sqrt{2} - 1)$$

Look at this problem [1] and other related questions to see the connection between Riemann Sums and Integrals.

[1] Calculate limit using Riemann integral

Solution 2:

Hint:

Just note \begin{align*} \lim_{n\to\infty}\left(\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+\ldots+\frac{1}{\sqrt{n}\sqrt{n+n}}\right)&=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}\frac{1}{n} \end{align*} Where $$\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n}}}\frac{1}{n}$$ is a Riemann sum for $f:[0,1]\to\mathbb{R}$ defined as $f(x)=\frac{1}{\sqrt{1+x}}$.

Solution 3:

First consider some algebraic manipulations and then use the integral idea appropriately: \begin{align} I &= \lim_{n\to\infty}\left(\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+\cdots+\frac{1}{\sqrt{n}\sqrt{n+n}}\right)\\[1em] &= \lim_{n\to\infty}\frac{1}{n}\left(\sqrt{\frac{n}{n+1}}+\sqrt{\frac{n}{n+2}}+\cdots+\sqrt{\frac{n}{n+n}}\right)\\[1em] &= \lim_{n\to\infty}\frac{1}{n}\left(\frac{1}{\sqrt{1+1/n}}+\frac{1}{\sqrt{1+2/n}}+\cdots+\frac{1}{\sqrt{1+1}}\right)\\[1em] &= \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^nf\left(\frac{i}{n}\right)\qquad\left[\text{where }f(x)=\frac{1}{\sqrt{1+x}}\right]\\[1em] &= \int_0^1 \frac{1}{\sqrt{1+x}}\,dx\\[1em] &= [2\sqrt{1+x}]\biggr\rvert_0^1\\[1em] &= 2(\sqrt{2}-1). \end{align}