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Question Number 60198 by Mr X pcx last updated on 18/May/19
$${find}\:{lim}_{{n}\rightarrow+\infty} \:{ln}\left(\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)^{\frac{\mathrm{1}}{{n}}} \right) \\ $$
Answered by tanmay last updated on 18/May/19
![=lim_(n→∞) ln[(1+(1^4 /n^4 ))(1+(2^4 /n^4 ))(1+(3^4 /n^4 ))...(1+(n^4 /n^4 ))]^(1/n) =lim_(n→∞) (1/n)×[ln(1+(1^4 /n^4 ))+ln(1+(2^4 /n^4 ))+..+ln(1+(n^4 /n^4 ))] h=(1/n) when n→∞ h→0 =lim_(h→0) h[ln(1+1^4 h^4 )+ln(1+2^4 h^4 )+..+ln(1+n^4 h^4 ) lim_(h→0) hΣ_(r=1) ^n ln(1+(rh)^4 ) ∫_0 ^1 ln(1+x^4 )dx now ∫ln(1+x^4 )dx =ln(1+x^4 )×x−∫(1/(1+x^4 ))×4x^3 ×xdx =xln(1+x^4 )−4∫((1+x^4 −1)/(1+x^4 ))dx =xln(1+x^4 )−4∫dx+4∫(dx/(1+x^4 )) =xln(1+x^4 )−4x+2∫((2/x^2 )/(x^2 +(1/x^2 )))dx =xln(1+x^4 )−4x+2∫(((1+(1/x^2 ))−(1−(1/x^2 )))/(x^2 +(1/x^2 )))dx =xln(1+x^4 )−4x+2∫((d(x−(1/x)))/((x−(1/x))^2 +2))−2∫((d(x+(1/x)))/((x+(1/x))^2 −2)) =xln(1+x^4 )−4x+2×(1/(√2))tan^(−1) (((x−(1/x))/(√2)))−2×(1/(2(√2)))ln(((x+(1/x)−(√2))/(x+(1/x)+(√2)))) so answer is =∣xln(1+x^4 )−4x+(√2) tan^(−1) (((x−(1/x))/(√2)))−(1/(√2))ln(((x^2 +1−(√2) x)/(x^2 +1+(√2) x)))∣_0 ^1 ={ln2−4+(√2) tan^−^1 (0)−(1/(√2))ln(((2−(√2))/(2+(√2))))}−{(√2) tan^(−1) (−∞)} =ln2−4−(1/(√2))ln(((2−(√2))/(2+(√2))))−(√2) ×(((−π)/2)) ln2−4−(1/(√2))ln(((2−(√2))/(2+(√2))))+(π/(√2)) pls check mistake if any...lengthy problem](Q60205.png)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{ln}\left[\left(\mathrm{1}+\frac{\mathrm{1}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)\left(\mathrm{1}+\frac{\mathrm{2}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)\left(\mathrm{1}+\frac{\mathrm{3}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)...\left(\mathrm{1}+\frac{{n}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)\right]^{\frac{\mathrm{1}}{{n}}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}×\left[{ln}\left(\mathrm{1}+\frac{\mathrm{1}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{2}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)+..+{ln}\left(\mathrm{1}+\frac{{n}^{\mathrm{4}} }{{n}^{\mathrm{4}} }\right)\right] \\ $$$${h}=\frac{\mathrm{1}}{{n}} \\ $$$${when}\:{n}\rightarrow\infty\:\:\:{h}\rightarrow\mathrm{0} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}{h}\left[{ln}\left(\mathrm{1}+\mathrm{1}^{\mathrm{4}} {h}^{\mathrm{4}} \right)+{ln}\left(\mathrm{1}+\mathrm{2}^{\mathrm{4}} {h}^{\mathrm{4}} \right)+..+{ln}\left(\mathrm{1}+{n}^{\mathrm{4}} {h}^{\mathrm{4}} \right)\right. \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}{h}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{1}+\left({rh}\right)^{\mathrm{4}} \right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right){dx} \\ $$$${now} \\ $$$$\int{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right){dx} \\ $$$$={ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)×{x}−\int\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }×\mathrm{4}{x}^{\mathrm{3}} ×{xdx} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{4}\int\frac{\mathrm{1}+{x}^{\mathrm{4}} −\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{4}\int{dx}+\mathrm{4}\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{4}{x}+\mathrm{2}\int\frac{\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{4}{x}+\mathrm{2}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{4}{x}+\mathrm{2}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}−\mathrm{2}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{4}{x}+\mathrm{2}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\sqrt{\mathrm{2}}}\right)−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}}\right) \\ $$$${so}\:{answer}\:{is} \\ $$$$=\mid{xln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{4}{x}+\sqrt{\mathrm{2}}\:{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{2}}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}}\:{x}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left\{{ln}\mathrm{2}−\mathrm{4}+\sqrt{\mathrm{2}}\:{tan}^{−^{\mathrm{1}} } \left(\mathrm{0}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)\right\}−\left\{\sqrt{\mathrm{2}}\:{tan}^{−\mathrm{1}} \left(−\infty\right)\right\} \\ $$$$={ln}\mathrm{2}−\mathrm{4}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)−\sqrt{\mathrm{2}}\:×\left(\frac{−\pi}{\mathrm{2}}\right) \\ $$$${ln}\mathrm{2}−\mathrm{4}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)+\frac{\pi}{\sqrt{\mathrm{2}}} \\ $$$${pls}\:{check}\:{mistake}\:{if}\:{any}...{lengthy}\:{problem} \\ $$