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let-A-n-k-1-n-1-k-2-n-2-calculate-lim-n-ln-A-n-n-




Question Number 66172 by mathmax by abdo last updated on 10/Aug/19
  let A_n =Π_(k=1) ^n (1+(k^2 /n^2 ))   calculate lim_(n→+∞)  ((ln(A_n ))/n)
letAn=k=1n(1+k2n2)calculatelimn+ln(An)n
Commented by Prithwish sen last updated on 11/Aug/19
lim_(n→∞) (1/n) Σ_(k=1) ^n ln[1+((k/n))^2 ] = ∫_0 ^1 ln(1+x^2 )dx  =[xln(1+x^2 )−2x+2tan^(−1) x]_0 ^1 =ln2 −2+(π/2)  please check.
limn1nnk=1ln[1+(kn)2]=01ln(1+x2)dx=[xln(1+x2)2x+2tan1x]01=ln22+π2pleasecheck.
Commented by mathmax by abdo last updated on 10/Aug/19
we have ln(A_n )=Σ_(k=1) ^n ln(1+(k^2 /n^2 )) ⇒((ln(A_n ))/n) =(1/n)Σ_(k=1) ^n ln(1+((k/n))^2 )  we get a Rieman sum ⇒lim_(n→+∞) ((ln(A_n ))/n) =∫_0 ^1 ln(1+x^2 )dx  by parts ∫_0 ^1 ln(1+x^2 )dx =[xln(1+x^2 )]_0 ^1  −∫_0 ^1 x((2x)/(1+x^2 ))dx  =ln(2)−2 ∫_0 ^1  ((x^2 +1−1)/(x^2  +1))dx =ln(2)−2 +2 ∫_0 ^1  (dx/(1+x^2 ))  =ln(2)−2 +2[arctanx]_0 ^1 =ln(2)−2+2×(π/4) =ln(2)−2+(π/2)  finally lim_(n→+∞)  ((ln(A_n ))/n) =(π/2) +ln(2)−2 .
wehaveln(An)=k=1nln(1+k2n2)ln(An)n=1nk=1nln(1+(kn)2)wegetaRiemansumlimn+ln(An)n=01ln(1+x2)dxbyparts01ln(1+x2)dx=[xln(1+x2)]0101x2x1+x2dx=ln(2)201x2+11x2+1dx=ln(2)2+201dx1+x2=ln(2)2+2[arctanx]01=ln(2)2+2×π4=ln(2)2+π2finallylimn+ln(An)n=π2+ln(2)2.
Commented by Prithwish sen last updated on 10/Aug/19
thanks sir.
thankssir.
Commented by mathmax by abdo last updated on 10/Aug/19
you are welcome sir.
youarewelcomesir.

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