Answer to Question #272455 in Real Analysis for Surbhi

Question #272455

Evaluate:



Limit x tends to infinity( n/1+n^2 + n/4+n^2 + n/9+n^2 +.......n/2n^2 )

1
Expert's answer
2021-11-30T13:43:28-0500

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 } }" .


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