calculate S(x) = Σ_(n=0) ^∞ (x^(3n) /(3n+1)) after finding the radius of convergence . 2) find the value of Σ


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Question Number 35621 by abdo mathsup 649 cc last updated on 21/May/18

$c a l c u l a t e S (x) = \sum_{n = 0}^{\infty} \frac{x^{3 n}}{3 n + 1} a f t e r f i n d i n g$ $t h e r a d i u s o f c o n v e r g e n c e .$ $2) f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{1}{(3 n + 1) 8^{n}}$
Commented by abdo.msup.com last updated on 25/May/18
$1) let u_n (x)= (x^(3n) /(3n+1)) if x≠0 ∣((u_(n+1) (x))/(u_n (x)))∣ = ∣ ((x^(3n+3) /(3n+4))/(x^(3n) /(3n+1)))∣ =((3n+1)/(3n+4)) ∣x∣^3 →∣x∣^3 so if ∣x∣<1 the serie converged so R =1 let find S(x) we have x S(x) = Σ_(n=0) ^∞ (x^(3n+1) /(3n+1)) =w(x) w^′ (x)= Σ_(n=0) ^∞ x^(3n) = (1/(1−x^3 )) ⇒ w(x) = ∫_0 ^x (dt/(1−t^3 )) +c but c=w(0)=0 F(t) = (1/(1−t^3 )) = (1/((1−t)(t^2 +t +1))) =(a/(1−t)) +((bt +c)/(t^2 +t +1)) a=lim_(t→1) (1−t)F(t) =(1/3) F(t) = (1/(3(1−t))) +((bt +c)/(t^2 +t +1)) lim_(t→+∞) t F(t) = −(1/3) +b =0⇒b=(1/3) F(t) = (1/(3(1−t))) +(((1/3)t +c)/(t^2 +t+1)) F(0) = (1/3) +c =1⇒c=1−(1/3) =(2/3) so F(t)= (1/(3(1−t))) +(1/3) ((t+2)/(t^2 +t+1)) 3w(x)=∫_0 ^x (dt/(1−t)) + (1/2)∫_0 ^x ((2t+1+3)/(t^2 +t+1))dt =[−ln∣1−t∣]_0 ^x +(1/2)[ln(t^2 +t+1)]_0 ^x +(3/2) ∫_0 ^x (dt/(t^2 +t+1)) =−ln∣1−x∣ +(1/2)ln(x^(2 ) +x+1)+(3/2) I I = ∫_0 ^x (dt/((t+(1/2))^2 +(3/4))) =_(t +(1/2)=((√3)/2) u) ∫_(1/(√3)) ^((2x+1)/(√3)) (1/((3/4)( 1+u^2 )))((√3)/2)du = (4/3) ((√3)/2) [arctan(u)]_(1/(√3)) ^((2x+1)/(√3)) = (2/(√3)) { arctan(((2x+1)/(√3))) −arctan((1/(√3)))} 3w(x)=−ln∣1−x∣ +(1/2)ln(x^2 +x+1) +(√3){arctan(((2x+1)/(√3))) −(π/6)} ⇒ w(x) =−(1/3)ln∣1−x∣ +(1/6)ln(x^2 +x+1) + (1/(√3)){ arctan(((2x+1)/(√3))) −(π/6)}and S(x)= ((w(x))/x)$
$1) l e t u_{n} (x) = \frac{x^{3 n}}{3 n + 1} i f x \neq 0$ $∣ \frac{u_{n + 1} (x)}{u_{n} (x)} ∣ = ∣ \frac{\frac{x^{3 n + 3}}{3 n + 4}}{\frac{x^{3 n}}{3 n + 1}} ∣ = \frac{3 n + 1}{3 n + 4} ∣ x ∣^{3}$ $\to∣ x ∣^{3} s o i f ∣ x ∣< 1 t h e s e r i e c o n v e r g e d$ $s o R = 1 l e t f i n d S (x)$ $w e h a v e x S (x) = \sum_{n = 0}^{\infty} \frac{x^{3 n + 1}}{3 n + 1} = w (x)$ $w^{^{'}} (x) = \sum_{n = 0}^{\infty} x^{3 n} = \frac{1}{1 - x^{3}} \Rightarrow$ $w (x) = \int_{0}^{x} \frac{d t}{1 - t^{3}} + c b u t c = w (0) = 0$ $F (t) = \frac{1}{1 - t^{3}} = \frac{1}{(1 - t) (t^{2} + t + 1)}$ $= \frac{a}{1 - t} + \frac{b t + c}{t^{2} + t + 1}$ $a = {l i m}_{t \to 1} (1 - t) F (t) = \frac{1}{3}$ $F (t) = \frac{1}{3 (1 - t)} + \frac{b t + c}{t^{2} + t + 1}$ ${l i m}_{t \to + \infty} t F (t) = - \frac{1}{3} + b = 0 \Rightarrow b = \frac{1}{3}$ $F (t) = \frac{1}{3 (1 - t)} + \frac{\frac{1}{3} t + c}{t^{2} + t + 1}$ $F (0) = \frac{1}{3} + c = 1 \Rightarrow c = 1 - \frac{1}{3} = \frac{2}{3}$ $s o F (t) = \frac{1}{3 (1 - t)} + \frac{1}{3} \frac{t + 2}{t^{2} + t + 1}$ $3 w (x) = \int_{0}^{x} \frac{d t}{1 - t} + \frac{1}{2} \int_{0}^{x} \frac{2 t + 1 + 3}{t^{2} + t + 1} d t$ $= {[- l n ∣ 1 - t ∣]}_{0}^{x} + \frac{1}{2} {[l n (t^{2} + t + 1)]}_{0}^{x}$ $+ \frac{3}{2} \int_{0}^{x} \frac{d t}{t^{2} + t + 1}$ $= - l n ∣ 1 - x ∣ + \frac{1}{2} l n (x^{2} + x + 1) + \frac{3}{2} I$ $I = \int_{0}^{x} \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}}$ $=_{t + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \int_{\frac{1}{\sqrt{3}}}^{\frac{2 x + 1}{\sqrt{3}}} \frac{1}{\frac{3}{4} (1 + u^{2})} \frac{\sqrt{3}}{2} d u$ $= \frac{4}{3} \frac{\sqrt{3}}{2} {[a r c t a n (u)]}_{\frac{1}{\sqrt{3}}}^{\frac{2 x + 1}{\sqrt{3}}}$ $= \frac{2}{\sqrt{3}} {a r c t a n (\frac{2 x + 1}{\sqrt{3}}) - a r c t a n (\frac{1}{\sqrt{3}})}$ $3 w (x) = - l n ∣ 1 - x ∣ + \frac{1}{2} l n (x^{2} + x + 1)$ $+ \sqrt{3} {a r c t a n (\frac{2 x + 1}{\sqrt{3}}) - \frac{π}{6}} \Rightarrow$ $w (x) = - \frac{1}{3} l n ∣ 1 - x ∣ + \frac{1}{6} l n (x^{2} + x + 1)$ $+ \frac{1}{\sqrt{3}} {a r c t a n (\frac{2 x + 1}{\sqrt{3}}) - \frac{π}{6}} a n d$ $S (x) = \frac{w (x)}{x}$
Commented by abdo.msup.com last updated on 25/May/18
$2) Σ_(n=0) ^∞ (1/((3n+1)8^n )) =S((1/2)) =2w((1/2))=2{ −(1/3)ln((1/2))+(1/6)ln((7/4)) +(1/(√3)){ arctan((2/(√3))) −(π/6)} =(2/3)ln(2) + (1/3)ln((7/4)) +(2/(√3))arctan((2/(√3))) −(π/(3(√3))) .$
$2) \sum_{n = 0}^{\infty} \frac{1}{(3 n + 1) 8^{n}} = S (\frac{1}{2})$ $= 2 w (\frac{1}{2}) = 2 {- \frac{1}{3} l n (\frac{1}{2}) + \frac{1}{6} l n (\frac{7}{4})$ $+ \frac{1}{\sqrt{3}} {a r c t a n (\frac{2}{\sqrt{3}}) - \frac{π}{6}}$ $= \frac{2}{3} l n (2) + \frac{1}{3} l n (\frac{7}{4}) + \frac{2}{\sqrt{3}} a r c t a n (\frac{2}{\sqrt{3}})$ $- \frac{π}{3 \sqrt{3}} .$
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