Answer to Question #181759 in Real Analysis for TUHIN SUBHRA DAS

Question

evaluate limit n tends to infinity [n/(1+n^2) + n/(4+n^2) + n/(9+n^2) +.....+n/2n^2

Answer 1

ANSWER: $\frac { \pi }{ 4 }$

Explanation:

$\frac { n }{ 1+\quad { n }^{ 2 } } +\frac { n }{ 4+{ n }^{ 2 } } +\frac { n }{ 9+{ n }^{ 2 } } +\cdots +\frac { n }{ \quad { n }^{ 2 }+{ n }^{ 2 } }$$=\frac { 1 }{ n } \left( \frac { 1 }{ \frac { 1 }{ { n }^{ 2 } } +\ 1 } +\frac { 1 }{ \frac { \ 4 }{ { n }^{ 2 } } +\ 1 } \quad +\frac { 1 }{ \frac { 9 }{ { n }^{ 2 } } +\ 1 } +\cdots +\frac { 1 }{ \frac { \ { n }^{ 2 }\ }{ { n }^{ 2 } } +\ 1 } \right) =$ $\sum _{ k=1 }^{ n }{ \frac { 1 }{ n } \cdot } \frac { 1 }{ { \quad \left( \frac { k }{ n } \right) }^{ 2 }+1 }$ .

For the function $f(x)=\frac { 1 }{ 1+{ x }^{ 2 } }$ this sum is the the integral Riemann sum on the interval [0,1]. Those sum $\sum _{ k=1 }^{ n }{ f({ c }_{ k })({ x }_{ k } } -{ x }_{ k-1 })$ , where ${ x }_{ 0 }=0,\quad { x }_{ k }={ c }_{ k }=\frac { k }{ n } ,\ k=1,2,...,n$ . Since the function is continuous , it is integrable. Therefore, $\lim _{ n\rightarrow \infty }{ \left( \frac { n }{ 1+\quad { n }^{ 2 } } +\frac { n }{ 4+{ n }^{ 2 } } +\frac { n }{ 9+{ n }^{ 2 } } +\cdots +\frac { n }{ \quad { n }^{ 2 }+{ n }^{ 2 } } \right) }$ =

$=\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } =\ \arctan { 1-\arctan { 0=\frac { \pi }{ 4 } } \quad } }$ .