Question Number 48065 by maxmathsup by imad last updated on 18/Nov/18
$$\left.{let}\:{f}\:\:\:\::\:\:\right]\mathrm{0},\mathrm{1}\left[\:\:{contnue}\:{integrable}\:\:{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {f}\left({x}\right){dx}\right. \\ $$$${prove}\:{that}\:\Sigma\:{u}_{{n}} \:{cnverge}\:{and}\:{find}\:{its}\:{sum} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 23/Nov/18
$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} {u}_{{n}} =\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {f}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}\right)^{{n}} {f}\left({x}\right)\right){dx}\:\left({due}\:{to}\:{uniform}\:{convergence}\:{on}\:\left[\mathrm{0},\mathrm{1}\right]\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{f}\left({x}\right)}{\mathrm{1}+{x}}\:{dx}\:{so}\:\Sigma\:{u}_{{n}} \:{converges}\:{and}\:{its}\:{sum}\:{is}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{f}\left({x}\right)}{\mathrm{1}+{x}}{dx}. \\ $$