Question Number 39517 by math khazana by abdo last updated on 07/Jul/18
$${find}\:{radius}\:{of}\:\:{S}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${and}\:{calculate}\:{its}\:{sum} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:\:{and}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:\mathrm{2}^{{n}} }\:. \\ $$$$ \\ $$
Commented by abdo mathsup 649 cc last updated on 07/Jul/18
$${for}\:{x}\neq\mathrm{0}\:\:\mid\:\frac{{u}_{{n}+\mathrm{1}} \left({x}\right)}{{u}_{{n}} \left({x}\right)}\mid\:=\mid\:\frac{\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }}{\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }}\mid=\frac{{n}^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\mid{x}\mid\rightarrow\mid{x}\mid \\ $$$${so}\:{if}\:\mid{x}\mid<\mathrm{1}\:\:{the}\:{serie}\:{converges} \\ $$$${if}\:{x}=−\mathrm{1}\:\:\Sigma\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:{converges}\left({alternate}\:{serie}\right) \\ $$$${if}\:{x}=\mathrm{1}\:\:\Sigma\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{is}\:{a}\:{reiman}\:{serie}\:{convergent} \\ $$$${and}\:{R}=\mathrm{1} \\ $$$${we}\:{have}\:\:\frac{{dS}}{{dx}}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{nx}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}} \\ $$$$=−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:\:\left(\:\:{we}\:{suppose}\:{that}\:{x}\neq\mathrm{0}\right)\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\:−\int_{\mathrm{0}} ^{{x}} \:\:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:+{c} \\ $$$${c}={S}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:{S}\left({x}\right)=−\int_{\mathrm{0}} ^{{x}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 07/Jul/18
$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:={S}\left(\mathrm{1}\right)\:\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:\:{dt}\: \\ $$$${and}\:{we}\:{have}\:{always}\:{proved}\:{that}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$