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Question Number 48065 by maxmathsup by imad last updated on 18/Nov/18
![let f : ]0,1[ contnue integrable u_n =(−1)^n ∫_0 ^1 x^n f(x)dx prove that Σ u_n cnverge and find its sum](Q48065.png)
$$\left.{let}\:{f}\:\:\:\::\:\:\right]\mathrm{0},\mathrm{1}\left[\:\:{contnue}\:{integrable}\:\:{u}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {f}\left({x}\right){dx}\right. \\ $$$${prove}\:{that}\:\Sigma\:{u}_{{n}} \:{cnverge}\:{and}\:{find}\:{its}\:{sum} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 23/Nov/18
![we have Σ_(n=0) ^∞ u_n =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n f(x)dx =∫_0 ^1 (Σ_(n=0) ^∞ (−x)^n f(x))dx (due to uniform convergence on [0,1] =∫_0 ^1 ((f(x))/(1+x)) dx so Σ u_n converges and its sum is ∫_0 ^1 ((f(x))/(1+x))dx.](Q48442.png)
$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} {u}_{{n}} =\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {f}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}\right)^{{n}} {f}\left({x}\right)\right){dx}\:\left({due}\:{to}\:{uniform}\:{convergence}\:{on}\:\left[\mathrm{0},\mathrm{1}\right]\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{f}\left({x}\right)}{\mathrm{1}+{x}}\:{dx}\:{so}\:\Sigma\:{u}_{{n}} \:{converges}\:{and}\:{its}\:{sum}\:{is}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{f}\left({x}\right)}{\mathrm{1}+{x}}{dx}. \\ $$