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Question: \(\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r = 1}^{2n}\frac{r}{\sqrt{n^{2} + r^{2}}}\) equals...

lim⁡n→∞1n∑r=12nrn2+r2\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r = 1}^{2n}\frac{r}{\sqrt{n^{2} + r^{2}}} equals

A

1+51 + \sqrt{5}

B

−1+5- 1 + \sqrt{5}

C

−1+2- 1 + \sqrt{2}

D

1+21 + \sqrt{2}

Answer

−1+5- 1 + \sqrt{5}

Explanation

Solution

L=lim⁡n→∞∑r=12n1n.r/n1+(r/n)2L = \lim_{n \rightarrow \infty}\sum_{r = 1}^{2n}{}\frac{1}{n}.\frac{r/n}{\sqrt{1 + (r/n)^{2}}}=∫02x1+x2dx=5−1\int_{0}^{2}{}\frac{x}{\sqrt{1 + x^{2}}}dx = \sqrt{5} - 1.