Question Number 69566 by Ajao yinka last updated on 25/Sep/19 | ||
Answered by mind is power last updated on 25/Sep/19 | ||
$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sum_{{k}=\mathrm{0}} ^{{x}} \int_{{k}} ^{{k}+\mathrm{1}} \left({x}−{k}\right)^{{k}} {e}^{−{nk}} {dx} \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sum_{{k}=\mathrm{0}} ^{{x}} {e}^{−{nk}} \int_{{k}} ^{{k}+\mathrm{1}} \left({x}−{k}\right)^{{k}} {dx} \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sum_{{k}=\mathrm{0}} ^{{x}} {e}^{−{nk}} \left[\frac{\left({x}−{k}\right)^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sum_{{k}=\mathrm{0}} ^{{x}} {e}^{−{nk}} .\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${A}=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sum_{{k}=\mathrm{0}} ^{{x}} {e}^{−\left({k}+\mathrm{1}\right){n}} .\frac{{e}^{{n}} }{{k}+\mathrm{1}} \\ $$$$ \\ $$$${we}\:{have}\:−\mathrm{ln}\:\left(\mathrm{1}−{x}\right)=\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}} \\ $$$${A}={e}^{{n}} \underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left({e}^{−{n}} \right)^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}={e}^{{n}} ×−{ln}\left(\mathrm{1}−{e}^{−{n}} \right)={e}^{{n}} {ln}\left(\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}} }\right)={e}^{{n}} \mathrm{ln}\:\left(\frac{{e}^{{n}} }{{e}^{{n}} −\mathrm{1}}\right)={e}^{{n}} \left({n}−{ln}\left({e}^{{n}} −\mathrm{1}\right)\right) \\ $$$$ \\ $$ | ||
Commented by Ajao yinka last updated on 26/Sep/19 | ||
$${perfect}\:{solution} \\ $$ | ||