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Question-193408

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Question Number 193408 by cortano12 last updated on 13/Jun/23
$$\:\underline{\underbrace{ }} \\ $$
Answered by MM42 last updated on 13/Jun/23
lim_(n→∞) (1/n)×(((1−(e^(a/n) )^n ×(1/e^(a/n) ))/(1−e^(a/n) )))= lim_(n→∞) (1/n)×(((e^(a/n) −(e^(a/n) )^n )/(1−e^(a/n) )))×(1/e^(a/n) )= lim_(n→∞) (1/n)×(1/e^(a/n) )×(((e^(a/n) −1+1−(e^(a/n) )^n )/(1−e^(a/n) )))= lim_(n→∞) (1/n)×(1/e^(a/n) )(((e^(a/n) −1)/(1−e^(a/n) )))+lim_(n→∞) (1/n)×(1/e^(a/n) )×(((1−(e^(a/n) )^n )/(1−e^(a/n) )))= lim_(n→∞) (((1/n)/(1−e^(a/n) )))=lim_(n→∞) (((−(1/n^2 ))/( (a/n^2 )e^(a/n) )) )=−(1/a) =lim_(n→∞) (1/n)×(1/e^(a/n) )×(((1−(e^(a/n) )^n )/(1−e^(a/n) )))= lim_(n→∞) (1/n)(1+e^(a/n) +e^((2a)/n) +...+e^((na)/n) )= lim_(n→∞) Σ_(i=1) ^n e^((i/n)a) ×(1/n) =∫_0 ^1 e^(ax) dx=((e^a −1)/a) ⇒ans=((e^a −2)/a) ✓
$${lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}×\left(\frac{\mathrm{1}−\left({e}^{\frac{{a}}{{n}}} \right)^{{n}} ×\frac{\mathrm{1}}{{e}^{\frac{{a}}{{n}}} }}{\mathrm{1}−{e}^{\frac{{a}}{{n}}} }\right)= \\ $$$${lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}×\left(\frac{{e}^{\frac{{a}}{{n}}} −\left({e}^{\frac{{a}}{{n}}} \right)^{{n}} }{\mathrm{1}−{e}^{\frac{{a}}{{n}}} }\right)×\frac{\mathrm{1}}{{e}^{\frac{{a}}{{n}}} }= \\ $$$${lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{{e}^{\frac{{a}}{{n}}} }×\left(\frac{{e}^{\frac{{a}}{{n}}} −\mathrm{1}+\mathrm{1}−\left({e}^{\frac{{a}}{{n}}} \right)^{{n}} }{\mathrm{1}−{e}^{\frac{{a}}{{n}}} }\right)= \\ $$$${lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{{e}^{\frac{{a}}{{n}}} }\left(\frac{{e}^{\frac{{a}}{{n}}} −\mathrm{1}}{\mathrm{1}−{e}^{\frac{{a}}{{n}}} }\right)+{lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{{e}^{\frac{{a}}{{n}}} }×\left(\frac{\mathrm{1}−\left({e}^{\frac{{a}}{{n}}} \right)^{{n}} }{\mathrm{1}−{e}^{\frac{{a}}{{n}}} }\right)= \\ $$$${lim}_{{n}\rightarrow\infty} \:\left(\frac{\frac{\mathrm{1}}{{n}}}{\mathrm{1}−{e}^{\frac{{a}}{{n}}} }\right)={lim}_{{n}\rightarrow\infty} \:\left(\frac{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}{\:\frac{{a}}{{n}^{\mathrm{2}} }{e}^{\frac{{a}}{{n}}} }\:\right)=−\frac{\mathrm{1}}{{a}} \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{{e}^{\frac{{a}}{{n}}} }×\left(\frac{\mathrm{1}−\left({e}^{\frac{{a}}{{n}}} \right)^{{n}} }{\mathrm{1}−{e}^{\frac{{a}}{{n}}} }\right)= \\ $$$${lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\left(\mathrm{1}+{e}^{\frac{{a}}{{n}}} +{e}^{\frac{\mathrm{2}{a}}{{n}}} +…+{e}^{\frac{{na}}{{n}}} \right)= \\ $$$${lim}_{{n}\rightarrow\infty} \:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:{e}^{\frac{{i}}{{n}}{a}} \:×\frac{\mathrm{1}}{{n}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{{ax}} {dx}=\frac{{e}^{{a}} −\mathrm{1}}{{a}}\: \\ $$$$\Rightarrow{ans}=\frac{{e}^{{a}} −\mathrm{2}}{{a}}\:\checkmark \\ $$

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