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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

provethat0⩽∫0∞t2e−ntet−1dt⩽1n2fornintegrnot0

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

∫0+∞t2e−ntet−1⩽1n2=∫0+∞t2e−(n+1)t1−e−tdt=∫0+∞t2.∑k⩾0e−t(k+n+1)tdt=∑k⩾0∫0+∞t2e−(k+n+1)tdt=∑k⩾0∫0+∞u2e−t(k+n+1)3dt=∑k⩾02(k+n+1)2claim(k+n+1)3⩾(k+n+1)2(k+n)2.2(2k+2n+1)proff⇔(k+n+1)(2k+2n+12)⩾(k+n)2clear⇒∀k∈N,∀n∈N∗(k+n+1)3⩾(k+n+1)2(k+n)2.22k+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⇒∑k⩾02(k+n+1)3⩽∑k⩾01(k+n)2−1(k+n+1)2=1n2⇒∫0+∞t2e−ntet−1dt<1n2t2e−ntet−1⩾0⇒∫0+∞t2e−ntet−1dt>0

Commented by abdomathmax last updated on 01/Dec/19

thank you sir.

thankyousir.

Commented by mind is power last updated on 02/Dec/19

y′re welcom

y′rewelcom

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