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Question Number 200300 by Calculusboy last updated on 16/Nov/23

Answered by MM42 last updated on 17/Nov/23

lnA=lim_(n→∞) (1/n) [ln(1+((1/n))^1 )+ln(1+((2/n))^2 +...+ln(1+((n/n))^2 ) =lim_(n→∞) (1/n) Σ_(i=1) ^n ln(1+((i/n))^2 ) =∫_0 ^1 ln(1+x^2 )dx =xln(1+x^2 )]_0 ^1 −2∫_0 ^1 (x^2 /(1+x^2 ))dx =ln2−2(x−tan^(−1) x )]_0 ^1 =ln2−2(1−(π/4)) ⇒A=2×(e)^(1/(π−4))

lnA=limn→∞1n[ln(1+(1n)1)+ln(1+(2n)2+...+ln(1+(nn)2)=limn→∞1n∑ni=1ln(1+(in)2)=∫01ln(1+x2)dx=xln(1+x2)]01−2∫01x21+x2dx=ln2−2(x−tan−1x)]01=ln2−2(1−π4)⇒A=2×eπ−4

Commented by Calculusboy last updated on 28/Nov/23

thanks sir

thankssir

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