Question Number 62420 by mathsolverby Abdo last updated on 20/Jun/19 | ||
$${let}\:{u}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{{n}^{{x}} }\:−\int_{{n}} ^{{n}+\mathrm{1}} \frac{{dt}}{{t}^{{x}} }\:\:{with}\:{x}\in\left[\mathrm{1},\mathrm{2}\right] \\ $$ $$\left.\mathrm{1}\right){prove}\:{that}\:\mathrm{0}\leqslant\:{u}_{{n}} \left({x}\right)\leqslant\frac{\mathrm{1}}{{n}^{{x}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{x}} }\:\left({n}>\mathrm{0}\right) \\ $$ $$\left.\mathrm{2}\right){prove}\:{that}\:\Sigma\:{u}_{{n}} \left({x}\right){converges} \\ $$ $${let}\:\gamma\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{u}_{{n}} \left(\mathrm{1}\right) \\ $$ $$\left.\mathrm{3}\right){find}\:\sum_{{n}=\mathrm{1}} ^{\infty} {u}_{{n}} \left({x}\right)\:{interms}\:{of}\:\xi\left({x}\right){and} \\ $$ $$\mathrm{1}−{x} \\ $$ $$\left.\mathrm{4}\right)\:{prove}\:{that}\:{the}\:{converg}.{of}\:\Sigma{u}_{{n}} \left({x}\right){is} \\ $$ $${uniform} \\ $$ $${prove}\:{that}\:{for}\:{x}\in{V}\left(\mathrm{1}\right) \\ $$ $$\xi\left({x}\right)\:=\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\gamma\:+{o}\left(\mathrm{1}\right) \\ $$ $$\left.\mathrm{5}\right)\:{find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{ln}\left({n}\right) \\ $$ | ||