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Question Number 66620 by Mohamed Amine Bouguezzoul last updated on 18/Aug/19

find lim_(n→∞) I_n I_n =∫_0 ^∞ (dx/((1+coth (nx))^n )) ,n≥1

$${find}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{I}_{{n}} \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+\mathrm{coth}\:\left({nx}\right)\right)^{{n}} }\:,{n}\geqslant\mathrm{1} \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 19/Aug/19

coth(nx) =((ch(nx))/(sh(nx))) =((e^(nx) +e^(−nx) )/(e^(nx) −e^(−nx) )) ⇒1+coth(nx) =1+((e^(nx) +e^(−nx) )/(e^(nx) −e^(−nx) )) =((e^(nx) −e^(−nx) +e^(nx) +e^(−nx) )/(e^(nx) −e^(−nx) )) =((2e^(nx) )/(e^(nx) (1−e^(−2nx) ))) =(2/(1−e^(−2nx) )) ⇒ I_n =∫_0 ^∞ (dx/(((2/(1−e^(−2nx) )))^n )) =∫_0 ^∞ (1/2^n )(1−e^(−nx) )^n dx =∫_0 ^∞ (1/2^n )e^(nln(1−e^(−nx) )) dx =∫_0 ^∞ f_n (x)dx with f_n (x) =(1/2^n )e^(nln(1−e^(−nx) )) we have ln(1−e^(−nx) )∼−e^(−nx) and nln(1−e^(−nx) )∼−ne^(−nx) ⇒ f_n (x)∼(1/2^n )e^(−ne^(−nx) ) →0 (n→+∞) the sequences (f_n )are continues lim_(n→+∞) I_n =∫_R^+ lim_(n→+∞) f_n (x) =0

$${coth}\left({nx}\right)\:=\frac{{ch}\left({nx}\right)}{{sh}\left({nx}\right)}\:=\frac{{e}^{{nx}} \:+{e}^{−{nx}} }{{e}^{{nx}} −{e}^{−{nx}} }\:\:\Rightarrow\mathrm{1}+{coth}\left({nx}\right)\:=\mathrm{1}+\frac{{e}^{{nx}} \:+{e}^{−{nx}} }{{e}^{{nx}} −{e}^{−{nx}} } \\ $$$$=\frac{{e}^{{nx}} −{e}^{−{nx}} \:+{e}^{{nx}} +{e}^{−{nx}} }{{e}^{{nx}} −{e}^{−{nx}} }\:=\frac{\mathrm{2}{e}^{{nx}} }{{e}^{{nx}} \left(\mathrm{1}−{e}^{−\mathrm{2}{nx}} \right)}\:=\frac{\mathrm{2}}{\mathrm{1}−{e}^{−\mathrm{2}{nx}} }\:\Rightarrow \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(\frac{\mathrm{2}}{\mathrm{1}−{e}^{−\mathrm{2}{nx}} }\right)^{{n}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\mathrm{1}−{e}^{−{nx}} \right)^{{n}} \:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{e}^{{nln}\left(\mathrm{1}−{e}^{−{nx}} \right)} {dx}\:\:=\int_{\mathrm{0}} ^{\infty} {f}_{{n}} \left({x}\right){dx}\:{with}\:{f}_{{n}} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{e}^{{nln}\left(\mathrm{1}−{e}^{−{nx}} \right)} \\ $$$${we}\:{have}\:{ln}\left(\mathrm{1}−{e}^{−{nx}} \right)\sim−{e}^{−{nx}} \:{and}\:{nln}\left(\mathrm{1}−{e}^{−{nx}} \right)\sim−{ne}^{−{nx}} \:\Rightarrow \\ $$$${f}_{{n}} \left({x}\right)\sim\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{e}^{−{ne}^{−{nx}} } \:\rightarrow\mathrm{0}\:\left({n}\rightarrow+\infty\right)\:\:\:{the}\:{sequences}\:\left({f}_{{n}} \right){are}\:{continues} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} =\int_{{R}^{+} } \:{lim}_{{n}\rightarrow+\infty} \:\:\:{f}_{{n}} \left({x}\right)\:=\mathrm{0} \\ $$

Commented by Mohamed Amine Bouguezzoul last updated on 20/Aug/19

excellent work abdo.

$${excellent}\:{work}\:{abdo}. \\ $$

Commented by mathmax by abdo last updated on 26/Aug/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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