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Question Number 35833 by abdo mathsup 649 cc last updated on 24/May/18
fond lim_(n→+∞) n^a {ln(1+e^(−n) ) −e^(−n) } with a>0
$${fond}\:{lim}_{{n}\rightarrow+\infty} \:\:{n}^{{a}} \:\:\left\{{ln}\left(\mathrm{1}+{e}^{−{n}} \right)\:−{e}^{−{n}} \right\}\:{with}\:{a}>\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 25/May/18
we have for ∣x∣<1 ln^′ (1+x) =(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ ln(1+x) =Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) so ln(1+x) = x −(x^2 /2) +o(x^3 )(x→0)⇒ ln(1 +e^(−n) ) = e^(−n) − (e^(−2n) /2) +o(e^(−3n) )(n→+∞) n^a { ln(1+e^(−n) )−e^(−n) } ∼ −(1/2)n^a e^(−2n) (n→+∞) but lim_(n→+∞) n^a e^(−2n) =lim_(n→+∞) e^(aln(n)−2n) =lim_(n→+∞) e^(n{ a((ln(n))/n) −2}) = lim_(n→+∞) e^(−2n) =0 so for all a>0 lim_(n→+∞) n^a {ln(1+e^(−n) ) −e^(−n) } =0
$${we}\:{have}\:{for}\:\mid{x}\mid<\mathrm{1}\:\:{ln}^{'} \left(\mathrm{1}+{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$$\Rightarrow\:{ln}\left(\mathrm{1}+{x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{x}^{{n}} }{{n}} \\ $$$${so}\:\:{ln}\left(\mathrm{1}+{x}\right)\:=\:{x}\:\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\left({x}\rightarrow\mathrm{0}\right)\Rightarrow \\ $$$${ln}\left(\mathrm{1}\:+{e}^{−{n}} \right)\:=\:{e}^{−{n}} \:−\:\frac{{e}^{−\mathrm{2}{n}} }{\mathrm{2}}\:\:+{o}\left({e}^{−\mathrm{3}{n}} \right)\left({n}\rightarrow+\infty\right) \\ $$$${n}^{{a}} \left\{\:{ln}\left(\mathrm{1}+{e}^{−{n}} \right)−{e}^{−{n}} \right\}\:\sim\:−\frac{\mathrm{1}}{\mathrm{2}}{n}^{{a}} \:{e}^{−\mathrm{2}{n}} \:\left({n}\rightarrow+\infty\right) \\ $$$${but}\:{lim}_{{n}\rightarrow+\infty} \:{n}^{{a}} \:{e}^{−\mathrm{2}{n}} \:={lim}_{{n}\rightarrow+\infty} {e}^{{aln}\left({n}\right)−\mathrm{2}{n}} \\ $$$$={lim}_{{n}\rightarrow+\infty} \:\:{e}^{{n}\left\{\:{a}\frac{{ln}\left({n}\right)}{{n}}\:−\mathrm{2}\right\}} \:=\:{lim}_{{n}\rightarrow+\infty} \:{e}^{−\mathrm{2}{n}} \:=\mathrm{0}\:{so} \\ $$$${for}\:{all}\:{a}>\mathrm{0}\:{lim}_{{n}\rightarrow+\infty} {n}^{{a}} \left\{{ln}\left(\mathrm{1}+{e}^{−{n}} \right)\:−{e}^{−{n}} \right\}\:=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18
when n→∞ then e^(−n) →0 so given limit (∞×0) form =((lim)/(n→∞))(({ln(1+e^(−n) )−e^(−n) })/(1/n^a )) ((0/0)) form using lhospitsl =((lim)/(n→∞))×(({(1/((1+e^(−n) )))×e^(−n) ×−1}−{e^(−n) ×(−1)})/(−a×n^(−a−1) )) =((lim)/(n→∞))×(({((−e^(−n) )/(1+e^(−n) ))}+e^(−n) )/(−a×n^(−a−1) )) =((lim)/(n→∞))×(({−e^(−n) +e^(−n) +e^(−2n) )/((1+e^(−n) )×(−a×n^(−a−1) ))) =(((−1)/a))×((lim)/(n→∞))(e^(−2n) /(1+e^(−n) ))×n^(a+1) =(((−1)/a))×((lim)/(n→∞))(1/(e^(2n) +e^n ))×n^(a+1) =(((−1)/a))×((lim)/(n→∞))(1/(e^n +1))×(n^(a+1) /e^n ) =((−/a))×((lim)/(n→∞))×((1/(e^(2n) +e^n ))/(1/n^(a+1) )) contd
$${when}\:{n}\rightarrow\infty\:\:{then}\:{e}^{−{n}} \rightarrow\mathrm{0} \\ $$$${so}\:{given}\:{limit}\:\left(\infty×\mathrm{0}\right)\:{form} \\ $$$$=\frac{{lim}}{{n}\rightarrow\infty}\frac{\left\{{ln}\left(\mathrm{1}+{e}^{−{n}} \right)−{e}^{−{n}} \right\}}{\frac{\mathrm{1}}{{n}^{{a}} }}\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right)\:{form} \\ $$$${using}\:{lhospitsl} \\ $$$$=\frac{{lim}}{{n}\rightarrow\infty}×\frac{\left\{\frac{\mathrm{1}}{\left(\mathrm{1}+{e}^{−{n}} \right)}×{e}^{−{n}} ×−\mathrm{1}\right\}−\left\{{e}^{−{n}} ×\left(−\mathrm{1}\right)\right\}}{−{a}×{n}^{−{a}−\mathrm{1}} } \\ $$$$=\frac{{lim}}{{n}\rightarrow\infty}×\frac{\left\{\frac{−{e}^{−{n}} }{\mathrm{1}+{e}^{−{n}} \:}\right\}+{e}^{−{n}} }{−{a}×{n}^{−{a}−\mathrm{1}} } \\ $$$$=\frac{{lim}}{{n}\rightarrow\infty}×\frac{\left\{−{e}^{−{n}} +{e}^{−{n}} +{e}^{−\mathrm{2}{n}} \right.}{\left(\mathrm{1}+{e}^{−{n}} \right)×\left(−{a}×{n}^{−{a}−\mathrm{1}} \right)} \\ $$$$=\left(\frac{−\mathrm{1}}{{a}}\right)×\frac{{lim}}{{n}\rightarrow\infty}\frac{{e}^{−\mathrm{2}{n}} }{\mathrm{1}+{e}^{−{n}} }×{n}^{{a}+\mathrm{1}} \\ $$$$=\left(\frac{−\mathrm{1}}{{a}}\right)×\frac{{lim}}{{n}\rightarrow\infty}\frac{\mathrm{1}}{{e}^{\mathrm{2}{n}} +{e}^{{n}} \:}×{n}^{{a}+\mathrm{1}} \\ $$$$=\left(\frac{−\mathrm{1}}{{a}}\right)×\frac{{lim}}{{n}\rightarrow\infty}\frac{\mathrm{1}}{{e}^{{n}} +\mathrm{1}}×\frac{{n}^{{a}+\mathrm{1}} }{{e}^{{n}} } \\ $$$$=\left(\frac{−}{{a}}\right)×\frac{{lim}}{{n}\rightarrow\infty}×\frac{\frac{\mathrm{1}}{{e}^{\mathrm{2}{n}} +{e}^{{n}} }}{\frac{\mathrm{1}}{{n}^{{a}+\mathrm{1}} }} \\ $$$${contd} \\ $$

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