Question Number 152601 by mnjuly1970 last updated on 30/Aug/21 | ||
$$ \\ $$$$\:\:\:{solve}.... \\ $$$$\:\:{lim}_{\:{n}\rightarrow\infty} \left\{\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}\:−\frac{{k}}{{n}}+\frac{{k}^{\:\mathrm{2}} }{{n}^{\:\mathrm{2}} }\:\right)^{\:\frac{\mathrm{1}}{{n}}} \right\}=? \\ $$$$\:\:{m}.{n}... \\ $$$$ \\ $$ | ||
Commented by mindispower last updated on 30/Aug/21 | ||
$${hello}\:{sir}\:{i}\:{can}'{tfind}\:{one}\:{post}\:{of}\:{you} \\ $$$$\int\frac{{sin}\left({x}\right)}{{x}^{{x}} }{dx}\:{somthig}\:{lik}\:{that} \\ $$$${have}\:{a}\:{good}\:{day}\:{god}\:{bless}\:{you} \\ $$ | ||
Answered by puissant last updated on 30/Aug/21 | ||
$${U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}−\frac{{k}}{{n}}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$ \\ $$$${V}_{{n}} ={ln}\left({U}_{{n}} \right) \\ $$$$\Rightarrow\:{lim}_{{n}\rightarrow\infty} {V}_{{n}} =\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{1}−\frac{{k}}{{n}}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right){dx} \\ $$$$\begin{cases}{{u}={ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}\\{{v}'=\mathrm{1}}\end{cases}\:\Rightarrow\:\begin{cases}{{u}'=\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}\\{{v}={x}}\end{cases} \\ $$$$=\left[{xln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −{x}}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }{dx} \\ $$$${To}\:{be}\:{continued}... \\ $$ | ||
Commented by SANOGO last updated on 30/Aug/21 | ||
$${merci}\:{bien}\:{mon}\:{prof} \\ $$ | ||
Commented by puissant last updated on 30/Aug/21 | ||