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Question Number 35621 by abdo mathsup 649 cc last updated on 21/May/18
$${calculate}\:\:\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{3}{n}} }{\mathrm{3}{n}+\mathrm{1}}\:\:{after}\:{finding} \\ $$$${the}\:{radius}\:{of}\:{convergence}\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\mathrm{8}^{{n}} } \\ $$
Commented by abdo.msup.com last updated on 25/May/18
![1) let u_n (x)= (x^(3n) /(3n+1)) if x≠0 ∣((u_(n+1) (x))/(u_n (x)))∣ = ∣ ((x^(3n+3) /(3n+4))/(x^(3n) /(3n+1)))∣ =((3n+1)/(3n+4)) ∣x∣^3 →∣x∣^3 so if ∣x∣<1 the serie converged so R =1 let find S(x) we have x S(x) = Σ_(n=0) ^∞ (x^(3n+1) /(3n+1)) =w(x) w^′ (x)= Σ_(n=0) ^∞ x^(3n) = (1/(1−x^3 )) ⇒ w(x) = ∫_0 ^x (dt/(1−t^3 )) +c but c=w(0)=0 F(t) = (1/(1−t^3 )) = (1/((1−t)(t^2 +t +1))) =(a/(1−t)) +((bt +c)/(t^2 +t +1)) a=lim_(t→1) (1−t)F(t) =(1/3) F(t) = (1/(3(1−t))) +((bt +c)/(t^2 +t +1)) lim_(t→+∞) t F(t) = −(1/3) +b =0⇒b=(1/3) F(t) = (1/(3(1−t))) +(((1/3)t +c)/(t^2 +t+1)) F(0) = (1/3) +c =1⇒c=1−(1/3) =(2/3) so F(t)= (1/(3(1−t))) +(1/3) ((t+2)/(t^2 +t+1)) 3w(x)=∫_0 ^x (dt/(1−t)) + (1/2)∫_0 ^x ((2t+1+3)/(t^2 +t+1))dt =[−ln∣1−t∣]_0 ^x +(1/2)[ln(t^2 +t+1)]_0 ^x +(3/2) ∫_0 ^x (dt/(t^2 +t+1)) =−ln∣1−x∣ +(1/2)ln(x^(2 ) +x+1)+(3/2) I I = ∫_0 ^x (dt/((t+(1/2))^2 +(3/4))) =_(t +(1/2)=((√3)/2) u) ∫_(1/(√3)) ^((2x+1)/(√3)) (1/((3/4)( 1+u^2 )))((√3)/2)du = (4/3) ((√3)/2) [arctan(u)]_(1/(√3)) ^((2x+1)/(√3)) = (2/(√3)) { arctan(((2x+1)/(√3))) −arctan((1/(√3)))} 3w(x)=−ln∣1−x∣ +(1/2)ln(x^2 +x+1) +(√3){arctan(((2x+1)/(√3))) −(π/6)} ⇒ w(x) =−(1/3)ln∣1−x∣ +(1/6)ln(x^2 +x+1) + (1/(√3)){ arctan(((2x+1)/(√3))) −(π/6)}and S(x)= ((w(x))/x)](Q35913.png)
$$\left.\mathrm{1}\right)\:{let}\:{u}_{{n}} \left({x}\right)=\:\frac{{x}^{\mathrm{3}{n}} }{\mathrm{3}{n}+\mathrm{1}}\:\:{if}\:{x}\neq\mathrm{0}\: \\ $$$$\mid\frac{{u}_{{n}+\mathrm{1}} \left({x}\right)}{{u}_{{n}} \left({x}\right)}\mid\:=\:\mid\:\frac{\frac{{x}^{\mathrm{3}{n}+\mathrm{3}} }{\mathrm{3}{n}+\mathrm{4}}}{\frac{{x}^{\mathrm{3}{n}} }{\mathrm{3}{n}+\mathrm{1}}}\mid\:=\frac{\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}{n}+\mathrm{4}}\:\mid{x}\mid^{\mathrm{3}} \\ $$$$\rightarrow\mid{x}\mid^{\mathrm{3}} \:\:\:{so}\:{if}\:\mid{x}\mid<\mathrm{1}\:\:{the}\:{serie}\:{converged} \\ $$$${so}\:{R}\:=\mathrm{1}\:{let}\:{find}\:{S}\left({x}\right) \\ $$$${we}\:{have}\:{x}\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{3}{n}+\mathrm{1}} }{\mathrm{3}{n}+\mathrm{1}}\:={w}\left({x}\right) \\ $$$${w}^{'} \left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{x}^{\mathrm{3}{n}} \:\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{3}} }\:\Rightarrow \\ $$$${w}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{3}} }\:\:+{c}\:\:\:{but}\:{c}={w}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{t}\right)\left({t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}\right)} \\ $$$$=\frac{{a}}{\mathrm{1}−{t}}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}} \\ $$$${a}={lim}_{{t}\rightarrow\mathrm{1}} \left(\mathrm{1}−{t}\right){F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−{t}\right)}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}} \\ $$$${lim}_{{t}\rightarrow+\infty} {t}\:{F}\left({t}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\:+{b}\:=\mathrm{0}\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−{t}\right)}\:+\frac{\frac{\mathrm{1}}{\mathrm{3}}{t}\:+{c}}{{t}^{\mathrm{2}} \:+{t}+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:+{c}\:=\mathrm{1}\Rightarrow{c}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${so}\:{F}\left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−{t}\right)}\:+\frac{\mathrm{1}}{\mathrm{3}}\:\frac{{t}+\mathrm{2}}{{t}^{\mathrm{2}} \:+{t}+\mathrm{1}} \\ $$$$\mathrm{3}{w}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\:\frac{{dt}}{\mathrm{1}−{t}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{2}{t}+\mathrm{1}+\mathrm{3}}{{t}^{\mathrm{2}} \:+{t}+\mathrm{1}}{dt}\: \\ $$$$=\left[−{ln}\mid\mathrm{1}−{t}\mid\right]_{\mathrm{0}} ^{{x}} \:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({t}^{\mathrm{2}} \:+{t}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{{x}} \\ $$$$+\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}+\mathrm{1}} \\ $$$$=−{ln}\mid\mathrm{1}−{x}\mid\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}\:} \:+{x}+\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{2}}\:{I} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=_{{t}\:+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{u}} \:\:\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{4}}\left(\:\mathrm{1}+{u}^{\mathrm{2}} \right)}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\left[{arctan}\left({u}\right)\right]_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}} \\ $$$$=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\left\{\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:−{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right\} \\ $$$$\mathrm{3}{w}\left({x}\right)=−{ln}\mid\mathrm{1}−{x}\mid\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right) \\ $$$$+\sqrt{\mathrm{3}}\left\{{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:−\frac{\pi}{\mathrm{6}}\right\}\:\Rightarrow \\ $$$${w}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{1}−{x}\mid\:+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right) \\ $$$$+\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:−\frac{\pi}{\mathrm{6}}\right\}{and} \\ $$$${S}\left({x}\right)=\:\frac{{w}\left({x}\right)}{{x}}\: \\ $$
Commented by abdo.msup.com last updated on 25/May/18
$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\mathrm{8}^{{n}} }\:={S}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{w}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\left\{\:−\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\frac{\mathrm{7}}{\mathrm{4}}\right)\right. \\ $$$$+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\right)\:−\frac{\pi}{\mathrm{6}}\right\} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{2}\right)\:\:+\:\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\frac{\mathrm{7}}{\mathrm{4}}\right)\:+\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\right) \\ $$$$−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$