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

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) 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) \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}} .