Question #181759

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


1
Expert's answer
2021-05-02T07:27:33-0400

ANSWER: π4\frac { \pi }{ 4 }

Explanation:

n1+n2+n4+n2+n9+n2++nn2+n2\frac { n }{ 1+\quad { n }^{ 2 } } +\frac { n }{ 4+{ n }^{ 2 } } +\frac { n }{ 9+{ n }^{ 2 } } +\cdots +\frac { n }{ \quad { n }^{ 2 }+{ n }^{ 2 } }=1n(11n2+ 1+1 4n2+ 1+19n2+ 1++1 n2 n2+ 1)==\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) = k=1n1n1(kn)2+1\sum _{ k=1 }^{ n }{ \frac { 1 }{ n } \cdot } \frac { 1 }{ { \quad \left( \frac { k }{ n } \right) }^{ 2 }+1 } .

For the function f(x)=11+x2f(x)=\frac { 1 }{ 1+{ x }^{ 2 } } this sum is the the integral Riemann sum on the interval [0,1]. Those sum k=1nf(ck)(xkxk1)\sum _{ k=1 }^{ n }{ f({ c }_{ k })({ x }_{ k } } -{ x }_{ k-1 }) , where x0=0,xk=ck=kn, k=1,2,...,n{ x }_{ 0 }=0,\quad { x }_{ k }={ c }_{ k }=\frac { k }{ n } ,\ k=1,2,...,n . Since the function is continuous , it is integrable. Therefore, limn(n1+n2+n4+n2+n9+n2++nn2+n2)\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) } =

=01dx1+x2= arctan1arctan0=π4=\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } =\ \arctan { 1-\arctan { 0=\frac { \pi }{ 4 } } \quad } } .


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