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Question Number 128499 by mathmax by abdo last updated on 07/Jan/21

find  u_n =∫_1 ^∞   (([ne^(−x) ])/n^3 )dx

$$\mathrm{find}\:\:\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{1}} ^{\infty} \:\:\frac{\left[\mathrm{ne}^{−\mathrm{x}} \right]}{\mathrm{n}^{\mathrm{3}} }\mathrm{dx} \\ $$

Answered by TheSupreme last updated on 08/Jan/21

m=ne^(−x)   x_m =log((n/m))  u_n =Σ_(i=1) ^n ∫_x_m  ^x_(m+1)  (m/n^3 )dx=  u_n =Σ_(m=1) ^n (m/n^3 )[log((n/(m+1)))−log((n/m))]  u_n =(1/n^3 )Σm log (m)−mlog(m+1)  u_n =(1/n^3 )Σ_(m=1) ^(⌊ne^(−1) ]) log(m+1)=  u_n =(1/n^3 )log({[ne^(−1) ]+1}!)

$${m}={ne}^{−{x}} \\ $$$${x}_{{m}} ={log}\left(\frac{{n}}{{m}}\right) \\ $$$${u}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\int_{{x}_{{m}} } ^{{x}_{{m}+\mathrm{1}} } \frac{{m}}{{n}^{\mathrm{3}} }{dx}= \\ $$$${u}_{{n}} =\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{m}}{{n}^{\mathrm{3}} }\left[{log}\left(\frac{{n}}{{m}+\mathrm{1}}\right)−{log}\left(\frac{{n}}{{m}}\right)\right] \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\Sigma{m}\:{log}\:\left({m}\right)−{mlog}\left({m}+\mathrm{1}\right) \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\underset{{m}=\mathrm{1}} {\overset{\left.\lfloor{ne}^{−\mathrm{1}} \right]} {\sum}}{log}\left({m}+\mathrm{1}\right)= \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{3}} }{log}\left(\left\{\left[{ne}^{−\mathrm{1}} \right]+\mathrm{1}\right\}!\right) \\ $$

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