Question
$$\mathop {\lim }\limits_{n\, \to \,\infty } \frac{1}{n}\sum\limits_{r\, = \,1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $$ equals:
A.
$$1 + \sqrt 5 $$
B.
$$ - 1 + \sqrt 5 $$
C.
$$ - 1 + \sqrt 2 $$
D.
$$1 + \sqrt 2 $$
Answer :
$$ - 1 + \sqrt 5 $$
Solution :
$$\eqalign{
& {\text{We have}} = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{1}{n}\sum\limits_{r\, = \,1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} \cr
& = \mathop {\lim }\limits_{n\, \to \,\infty } \frac{1}{n}\sum\limits_{r\, = \,1}^{2n} {\frac{r}{{n\sqrt {1 + {{\left( {\frac{r}{n}} \right)}^2}} }}} \cr
& = \int_0^2 {\frac{x}{{\sqrt {1 + {x^2}} }}dx} \,\,\,\left[ {\because \mathop {\lim }\limits_{n\, \to \,\infty } \frac{1}{r}\sum\limits_{r\, = \,0}^{{a_n}} {f\left( {\frac{r}{n}} \right) = \int\limits_0^a {f\left( x \right)dx} } } \right] \cr
& = \left[ {\sqrt {1 + {x^2}} } \right]_0^2 = \sqrt 5 - 1 \cr} $$