Mathematics

$$\displaystyle \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{2n}\frac{r}{\sqrt{n^{2}+r^{2}}}$$ equals


ANSWER

$$\displaystyle -1+\sqrt{5}$$


SOLUTION
$$\displaystyle \lim _{ n\rightarrow \infty  } \dfrac { 1 }{ n } \sum _{ r=1 }^{ 2n } \dfrac { r }{ \sqrt { n^{ 2 }+r^{ 2 } }  } =\lim _{ n\rightarrow \infty  } \dfrac { 1 }{ n } \sum _{ r=1 }^{ 2n } \dfrac { \dfrac { r }{ n }  }{ \sqrt { 1+\dfrac { r^{ 2 } }{ { n }^{ 2 } }  }  } $$

$$\displaystyle =\int _{ 0 }^{ 2 }{ \dfrac { x }{ \sqrt { 1+{ x }^{ 2 } }  } dx } ={ \left[ \sqrt { 1+{ x }^{ 2 } }  \right]  }_{ 0 }^{ 2 }=\sqrt { 5 } -1$$
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Single Correct Medium Published on 17th 09, 2020
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