# 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?

• Hint: This is a Riemann sum. – Lucian Feb 13 '15 at 4:23
• Are you aware of the relationship between $\sum$ and $\int$? – Daniel W. Farlow Feb 13 '15 at 4:23
• The terms go to zero but there are more of them as $n$ grows. – Ian Feb 13 '15 at 4:24
• Bring a $\frac{1}{\sqrt{n}}$ outside each $\frac{1}{\sqrt{n+k}}$ to join the $\sqrt{n}$ which is already outside. – André Nicolas Feb 13 '15 at 4:24

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