prove-that-0-0-t-2-e-nt-e-t-1-dt-1-n-2-for-n-integr-not-0-

Question Number 74799 by mathmax by abdo last updated on 30/Nov/19

p r o v e t h a t 0 ⩽ \int_{0}^{\infty} \frac{t^{2} e^{- n t}}{e^{t} - 1} d t ⩽ \frac{1}{n^{2}} f o r n i n t e g r n o t 0

Answered by mind is power last updated on 30/Nov/19

\int_{0}^{+ \infty} \frac{t^{2} e^{- nt}}{e^{t} - 1} ⩽ \frac{1}{n^{2}}

= \int_{0}^{+ \infty} \frac{t^{2} e^{- (n + 1) t}}{1 - e^{- t}} dt

= \int_{0}^{+ \infty} t^{2} . \sum_{k ⩾ 0} e^{- t (k + n + 1) t} dt

= \sum_{k ⩾ 0} \int_{0}^{+ \infty} t^{2} e^{- (k + n + 1) t} dt

= \sum_{k ⩾ 0} \int_{0}^{+ \infty} \frac{u^{2} e^{- t}}{{(k + n + 1)}^{3}} dt = \sum_{k ⩾ 0} \frac{2}{{(k + n + 1)}^{2}}

claim

{(k + n + 1)}^{3} ⩾ {(k + n + 1)}^{2} {(k + n)}^{2} . \frac{2}{(2 k + 2 n + 1)}

proff \Leftrightarrow (k + n + 1) (\frac{2 k + 2 n + 1}{2}) ⩾ {(k + n)}^{2} clear

\Rightarrow \forall k \in N, \forall n \in N^{*}

{(k + n + 1)}^{3} ⩾ {(k + n + 1)}^{2} {(k + n)}^{2} . \frac{2}{2 k + 2 n + 1}

\Leftrightarrow \frac{2}{{(k + n + 1)}^{3}} ⩽ \frac{2 k + 2 n + 1}{{(k + n + 1)}^{2} {(k + n)}^{2}} = \frac{{(k + n + 1)}^{2} - {(k + n)}^{2}}{{(k + n + 1)}^{2} {(k + n)}^{2}} = \frac{1}{{(k + n)}^{2}} - \frac{1}{{(k + n + 1)}^{2}}

\Rightarrow \sum_{k ⩾ 0} \frac{2}{{(k + n + 1)}^{3}} ⩽ \sum_{k ⩾ 0} \frac{1}{{(k + n)}^{2}} - \frac{1}{{(k + n + 1)}^{2}} = \frac{1}{n^{2}}

\Rightarrow \int_{0}^{+ \infty} \frac{t^{2} e^{- nt}}{e^{t} - 1} dt < \frac{1}{n^{2}}

\frac{t^{2} e^{- nt}}{e^{t} - 1} ⩾ 0 \Rightarrow \int_{0}^{+ \infty} t^{2} \frac{e^{- nt}}{e^{t} - 1} dt > 0

Commented by abdomathmax last updated on 01/Dec/19

t h a n k y o u s i r .

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

y^{'} re welcom