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Question Number 74799 by mathmax by abdo last updated on 30/Nov/19
prove that 0≤∫_0 ^∞ ((t^2 e^(−nt) )/(e^t −1))dt ≤(1/n^2 ) for n integr not 0
$${prove}\:{that}\:\mathrm{0}\leqslant\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} \:{e}^{−{nt}} }{{e}^{{t}} −\mathrm{1}}{dt}\:\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{for}\:{n}\:{integr}\:{not}\:\mathrm{0} \\ $$
Answered by mind is power last updated on 30/Nov/19
∫_0 ^(+∞) ((t^2 e^(−nt) )/(e^t −1))≤(1/n^2 ) =∫_0 ^(+∞) ((t^2 e^(−(n+1)t) )/(1−e^(−t) ))dt =∫_0 ^(+∞) t^2 .Σ_(k≥0) e^(−t(k+n+1)t) dt =Σ_(k≥0) ∫_0 ^(+∞) t^2 e^(−(k+n+1)t) dt =Σ_(k≥0) ∫_0 ^(+∞) ((u^2 e^(−t) )/((k+n+1)^3 ))dt=Σ_(k≥0) (2/((k+n+1)^2 )) claim (k+n+1)^3 ≥(k+n+1)^2 (k+n)^2 .(2/((2k+2n+1))) proff⇔(k+n+1)(((2k+2n+1)/2))≥(k+n)^2 clear ⇒∀k∈N,∀n∈N^∗ (k+n+1)^3 ≥(k+n+1)^2 (k+n)^2 .(2/(2k+2n+1)) ⇔(2/((k+n+1)^3 ))≤((2k+2n+1)/((k+n+1)^2 (k+n)^2 ))=(((k+n+1)^2 −(k+n)^2 )/((k+n+1)^2 (k+n)^2 ))=(1/((k+n)^2 ))−(1/((k+n+1)^2 )) ⇒Σ_(k≥0) (2/((k+n+1)^3 ))≤Σ_(k≥0) (1/((k+n)^2 ))−(1/((k+n+1)^2 ))=(1/n^2 ) ⇒∫_0 ^(+∞) ((t^2 e^(−nt) )/(e^t −1))dt<(1/n^2 ) ((t^2 e^(−nt) )/(e^t −1))≥0⇒∫_0 ^(+∞) t^2 (e^(−nt) /(e^t −1))dt>0
$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\mathrm{nt}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}\leqslant\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\left(\mathrm{n}+\mathrm{1}\right)\mathrm{t}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{t}} }\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \mathrm{t}^{\mathrm{2}} .\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\mathrm{e}^{−\mathrm{t}\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)\mathrm{t}} \mathrm{dt}\:\: \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} \mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)\mathrm{t}} \mathrm{dt} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{u}^{\mathrm{2}} \mathrm{e}^{−\mathrm{t}} }{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} \:}\mathrm{dt}=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}}{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{claim} \\ $$$$\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} \geqslant\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} .\frac{\mathrm{2}}{\left(\mathrm{2k}+\mathrm{2n}+\mathrm{1}\right)} \\ $$$$\mathrm{proff}\Leftrightarrow\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)\left(\frac{\mathrm{2k}+\mathrm{2n}+\mathrm{1}}{\mathrm{2}}\right)\geqslant\left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} \:\mathrm{clear} \\ $$$$\Rightarrow\forall\mathrm{k}\in\mathbb{N},\forall\mathrm{n}\in\mathbb{N}^{\ast} \\ $$$$\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} \geqslant\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} .\frac{\mathrm{2}}{\mathrm{2k}+\mathrm{2n}+\mathrm{1}} \\ $$$$\Leftrightarrow\frac{\mathrm{2}}{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\leqslant\frac{\mathrm{2k}+\mathrm{2n}+\mathrm{1}}{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} }=\frac{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} }{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}}{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\leqslant\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{n}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\mathrm{nt}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}\mathrm{dt}<\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\mathrm{nt}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}\geqslant\mathrm{0}\Rightarrow\int_{\mathrm{0}} ^{+\infty} \mathrm{t}^{\mathrm{2}} \frac{\mathrm{e}^{−\mathrm{nt}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}\mathrm{dt}>\mathrm{0} \\ $$$$ \\ $$
Commented by abdomathmax last updated on 01/Dec/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 02/Dec/19
y′re welcom
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$

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