Question Number 128499 by mathmax by abdo last updated on 07/Jan/21
![find u_n =∫_1 ^∞ (([ne^(−x) ])/n^3 )dx](Q128499.png)
$$\mathrm{find}\:\:\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{1}} ^{\infty} \:\:\frac{\left[\mathrm{ne}^{−\mathrm{x}} \right]}{\mathrm{n}^{\mathrm{3}} }\mathrm{dx} \\ $$
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}!)](Q128504.png)
$${m}={ne}^{−{x}} \\ $$$${x}_{{m}} ={log}\left(\frac{{n}}{{m}}\right) \\ $$$${u}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\int_{{x}_{{m}} } ^{{x}_{{m}+\mathrm{1}} } \frac{{m}}{{n}^{\mathrm{3}} }{dx}= \\ $$$${u}_{{n}} =\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{m}}{{n}^{\mathrm{3}} }\left[{log}\left(\frac{{n}}{{m}+\mathrm{1}}\right)−{log}\left(\frac{{n}}{{m}}\right)\right] \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\Sigma{m}\:{log}\:\left({m}\right)−{mlog}\left({m}+\mathrm{1}\right) \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\underset{{m}=\mathrm{1}} {\overset{\left.\lfloor{ne}^{−\mathrm{1}} \right]} {\sum}}{log}\left({m}+\mathrm{1}\right)= \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{3}} }{log}\left(\left\{\left[{ne}^{−\mathrm{1}} \right]+\mathrm{1}\right\}!\right) \\ $$