# Evaluate $\lim_{n \to \infty }\left(\frac{1}{{n\sqrt{{n^2} + 1}}}+\frac{2}{{n\sqrt{{n^2}+4}}}+\cdots+\frac{n}{{n\sqrt{{n^2}+{n^2}}}}\right)$ [closed]

Evaluate $$\mathop {\lim }\limits_{n \to \infty } \left(\frac{1}{{n\sqrt {{n^2} + 1} }} + \frac{2}{{n\sqrt {{n^2} + 4} }} + \frac{3}{{n\sqrt {{n^2} + 9} }} + \cdots + \frac{n}{{n\sqrt {{n^2} + {n^2}} }}\right).$$

## closed as off-topic by user21820, YuiTo Cheng, José Carlos Santos, mrtaurho, Lee David Chung LinMay 9 at 8:34

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• Hint: Riemann sums – user258700 Jul 7 '16 at 11:29

## 2 Answers

You need to apply the principle of Riemann sum.

\begin{align} \\ & \lim\limits_{n \to \infty } \left(\frac{1}{{n\sqrt {{n^2} + 1} }} + \frac{2}{{n\sqrt {{n^2} + 4} }} + \frac{3}{{n\sqrt {{n^2} + 9} }} + \cdots + \frac{n}{{n\sqrt {{n^2} + {n^2}} }}\right)\\ & = \lim\limits_{n \to \infty }\sum_{r=1}^n \frac{r}{n\sqrt{n^2+r^2}} \\ & = \lim\limits_{n \to \infty }\sum_{r=1}^n \frac{\frac{r}{n}}{n\sqrt{1+(\frac{r}{n})^2}}\\ & = \lim\limits_{n \to \infty } \frac{1}{n} \sum_{r=1}^n \frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^2}} \end{align}

Now assume $\frac1n=h$. Then $h\to 0$ as $n\to\infty$.

Hence we get \begin{align} \\ & \lim\limits_{n \to \infty } \frac{1}{n} \sum_{r=1}^n \frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^2}}\\ & = \lim\limits_{n \to \infty } h \sum_{r=1}^n \frac{rh}{\sqrt{1+(rh)^2}}\\ & = \int_0^1 \frac{x}{\sqrt{1+x^2}}dx \\ & = \frac{1}{2}\int_0^1 \frac{d(1+x^2)}{\sqrt{1+x^2}}dx \\ & = \frac{1}{2}\cdot 2\sqrt{1+x^2}|_0^1 \\ & = \color{red}{\sqrt2-1} \end{align}

$$\dfrac1n\sum_{r=1}^n\dfrac r{\sqrt{n^2+r^2}} =\dfrac1n\sum_{r=1}^n\dfrac{r/n}{\sqrt{1+(r/n)^2}}$$

Now use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ and set $\sqrt{1+x^2}=y$

Can you take it from here?