(100 root 3) days streak problem

Calculus Level 5

lim n 1 n 4 [ ( n 2 + 1 2 ) ( n 2 + 2 2 ) ( n 2 + 3 2 ) ( n 2 + 4 n 2 ) ] 1 n \large \displaystyle \lim_{n\to\infty} \frac1{n^4} \left[ (n^2+1^2)(n^2+2^2)(n^2+3^2)\ldots (n^2+4n^2) \right]^{\frac 1n}

If the limit above equals to a e b tan 1 ( c ) b c ae^{b \tan^{-1}(c) - bc} for integers a , b , c a,b,c with e = m = 0 1 m ! \displaystyle e =\sum_{m=0}^\infty \frac1{m!} , calculate a + b + c a+b+c .


The answer is 29.

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1 solution

Vaibhav Sahu
May 24, 2015

y = lim n 1 n 4 [ ( n 2 + 1 2 ) ( n 2 + 2 2 ) . . . . . . ( n 2 + 4 n 2 ) ] ln y = lim n r = 0 2 n ln ( n 2 + r 2 n 2 ) n ln y = lim n r = 0 2 n 2 2 n ln [ 1 + 4 ( r 2 n ) 2 ] \\ y=\lim _{ n\to \infty } \frac { 1 }{ { n }^{ 4 } } [({ n }^{ 2 }+{ 1 }^{ 2 })({ n }^{ 2 }+{ 2 }^{ 2 })......({ n }^{ 2 }+4{ n }^{ 2 })]\\ \ln { y } =\lim _{ n\to \infty } \frac { \sum _{ r=0 }^{ 2n }{ \ln { (\frac { { n }^{ 2 }+{ r }^{ 2 } }{ { n }^{ 2 } } } } ) }{ n } \\ \ln { y } =\lim _{ n\to \infty } \sum _{ r=0 }^{ 2n }{ \frac { 2 }{ 2n } \ln { [1+4{ (\frac { r }{ 2n } })^{ 2 }] } } \\ From the definition of integral :- ln y = 0 1 2 ln ( 1 + 4 x 2 ) d x ln y = 2 ln 5 + 2 tan 1 2 4 y = 25 e 2 ( tan 1 2 2 ) a = 25 , b = 2 , c = 2 a + b + c = 29 \ln { y } =\int _{ 0 }^{ 1 }{ 2\ln { (1+4{ x }^{ 2 }) } } dx\\ \ln { y } =2\ln { 5+2\tan ^{ -1 }{ 2 } -4 } \\ y=25{ e }^{ 2(\tan ^{ -1 }{ 2 } -2) }\\ a=25,\quad b=2,\quad c=2\\ a+b+c=29

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