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? 
 A: $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+\cdots+\frac{1}{\sqrt{n}\sqrt{n+n}}$$
$$=\lim_{n\rightarrow\infty}\sum_{i = 1}^n\frac{1}{\sqrt{n}\sqrt{n+i}} = \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i = 1}^n\frac{1}{\sqrt{1+i/n}} = \int_0^1 \frac{1}{\sqrt{1 + x}} = 2(\sqrt{2} - 1)$$
Look at this problem [1] and other related questions to see the connection between Riemann Sums and Integrals.
[1] Calculate limit using Riemann integral
A: 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 $f(x)=\frac{1}{\sqrt{1+x}}$.
A: 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}
