Menu Close

calculate-S-x-n-0-x-3n-3n-1-after-finding-the-radius-of-convergence-2-find-the-value-of-n-0-1-3n-1-8-n-




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

Leave a Reply

Your email address will not be published. Required fields are marked *