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Question Number 56708 by gunawan last updated on 22/Mar/19
for all n ∈ N, f_n (x)=Σ_(k=1) ^n (−1)^n x^n   for any −1<x<1. If f : (−1,1)→R  with f =lim_(n→∞)  f_(n ) of (−1,1)  ∫_0 ^(1/2) f(x) dx=...
$$\mathrm{for}\:\mathrm{all}\:{n}\:\in\:\mathbb{N},\:{f}_{{n}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$$\mathrm{for}\:\mathrm{any}\:−\mathrm{1}<{x}<\mathrm{1}.\:\mathrm{If}\:{f}\::\:\left(−\mathrm{1},\mathrm{1}\right)\rightarrow\mathbb{R} \\ $$$$\mathrm{with}\:{f}\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{f}_{{n}\:} \mathrm{of}\:\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right)\:{dx}=… \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 22/Mar/19
we have f_n (x)=Σ_(k=1) ^n (−x)^k  =Σ_(k=0) ^n (−x)^k −1  =((1−(−x)^(n+1) )/(1+x)) −1 =((1−(−x)^(n+1) −1−x)/(1+x))  but ∣x∣<1 ⇒lim_(n→+∞)    f_n (x)=((−x)/(1+x)) ⇒  f(x)=((−x)/(1+x)) ⇒∫_0 ^(1/2) f(x)dx =−∫_0 ^(1/2)   ((1+x−1)/(1+x))dx  =−(1/2) +∫_0 ^(1/2)  (dx/(1+x)) =−(1/2) +[ln∣1+x∣]_0 ^(1/2)   =−(1/2) +ln((3/2)) .  here i have used that lim f_n =f .
$${we}\:{have}\:{f}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \left(−{x}\right)^{{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{x}\right)^{{k}} −\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\left(−{x}\right)^{{n}+\mathrm{1}} }{\mathrm{1}+{x}}\:−\mathrm{1}\:=\frac{\mathrm{1}−\left(−{x}\right)^{{n}+\mathrm{1}} −\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$${but}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:\:{f}_{{n}} \left({x}\right)=\frac{−{x}}{\mathrm{1}+{x}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{−{x}}{\mathrm{1}+{x}}\:\Rightarrow\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right){dx}\:=−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{x}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:+\left[{ln}\mid\mathrm{1}+{x}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:+{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:. \\ $$$${here}\:{i}\:{have}\:{used}\:{that}\:{lim}\:{f}_{{n}} ={f}\:. \\ $$

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