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))](Q200310.png)
$${lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\left[{ln}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{1}} \right)+{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} +...+{ln}\left(\mathrm{1}+\left(\frac{{n}}{{n}}\right)^{\mathrm{2}} \right)\right.\right. \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{1}+\left(\frac{{i}}{{n}}\right)^{\mathrm{2}} \right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx} \\ $$$$ \\ $$$$\left.={xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\left.={ln}\mathrm{2}−\mathrm{2}\left({x}−{tan}^{−\mathrm{1}} {x}\:\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={ln}\mathrm{2}−\mathrm{2}\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow{A}=\mathrm{2}×\sqrt[{\pi−\mathrm{4}}]{{e}}\: \\ $$$$ \\ $$
Commented by Calculusboy last updated on 28/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$