find-the-radius-of-S-x-n-0-x-3n-2-3n-2-2-find-the-value-of-n-0-1-3n-2-3-n-

Question Number 29979 by abdo imad last updated on 14/Feb/18

f i n d t h e r a d i u s o f S (x) = \sum_{n = 0}^{\infty} \frac{x^{3 n + 2}}{3 n + 2}

2) f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{1}{(3 n + 2) 3^{n}} .

Commented by abdo imad last updated on 17/Feb/18

1) f o r x = 0 S (x) = 0 f o r x \neq o w e h a v e

∣ \frac{u_{n + 1} (x)}{u_{n} (x)} ∣=∣ \frac{\frac{x^{3 n + 5}}{3 n + 5}}{\frac{x^{3 n + 2}}{3 n + 2}} ∣= \frac{3 n + 2}{3 n + 5} ∣ x ∣_{n \to \infty}^{3} \to∣ x ∣^{3} \Rightarrow i f ∣ x ∣< 1

t h e s e r i e c o n v e r g e i f ∣ x ∣⩾ 1 t h e s e r i e d i v e r g e i f x = - 1

S (- 1) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 2} c o n v e r g e (a l t e r n a t e s e r i e) s o R = 1

2) f o r ∣ x ∣< 1 a n d d u e t o u n i f o r m c o n v e r g e n c e w e h a v e

S^{^{'}} (x) = \sum_{n = 0}^{\infty} x^{3 n + 1} = x \sum_{n = 0}^{\infty} {(x^{3})}^{n} = x \frac{1}{1 - x^{3}} \Rightarrow

S (x) = \int \frac{x d x}{1 - x^{3}} + λ l e t d e c o m p o s e

F (x) = \frac{x}{1 - x^{3}} = \frac{x}{(1 - x) (x^{2} + x + 1)} = \frac{a}{1 - x} + \frac{b x + c}{x^{2} + x + 1}

a = {l i m}_{x \to 1} (1 - x) F (x) = \frac{1}{3} s o

F (x) = \frac{1}{3 (1 - x)} + \frac{b x + c}{x^{2} + x + 1}

{l i m}_{x \to \infty} x F (x) = 0 = \frac{- 1}{3} + b \Rightarrow b = \frac{1}{3} \Rightarrow

F (x) = \frac{1}{3 (1 - x)} + \frac{\frac{1}{3} x + c}{x^{2} + x + 1}

F (0) = 0 = \frac{1}{3} + c \Rightarrow c = - \frac{1}{3} a n d

F (x) = \frac{1}{3 (1 - x)} + \frac{1}{3} \frac{x - 1}{x^{2} + x + 1} \Rightarrow

\int \frac{x}{1 - x^{3}} d x = \frac{1}{3} (\int \frac{d x}{1 - x} + \int \frac{x - 1}{x^{2} + x + 1} d x)

3 \int \frac{x}{1 - x^{3}} d x = \int \frac{d x}{1 - x} + \frac{1}{2} \int \frac{2 x + 1 - 3}{x^{2} + x + 1} d x

= \frac{1}{2} l n (x^{2} + x + 1) - l n ∣ 1 - x ∣ - \frac{3}{2} \int \frac{d x}{x^{2} + x + 1} b u t

\int \frac{d x}{x^{2} + x + 1} = \int \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int \frac{1}{\frac{3}{4} (t^{2} + 1)} \frac{\sqrt{3}}{2} d t

= \frac{4}{3} \frac{\sqrt{3}}{2} \int \frac{d t}{1 + t^{2}} = \frac{2}{\sqrt{3}} a r t a n t = \frac{2}{\sqrt{3}} a r c t a n (\frac{2}{\sqrt{3}} (x + \frac{1}{2}))

3 \int \frac{x}{1 - x^{2}} d x = l n (\frac{\sqrt{x^{2} + x + 1}}{∣ 1 - x ∣}) - \sqrt{3} a r c t a n (\frac{1}{\sqrt{3}} (2 x + 1))

\int \frac{x d x}{1 - x^{2}} = \frac{1}{3} l n (\frac{\sqrt{x^{2} + x + 1}}{∣ 1 - x ∣}) - \frac{\sqrt{3}}{3} a r c t a n (\frac{1}{\sqrt{3}} (2 x + 1))

S (x) = \frac{1}{3} l n (\frac{\sqrt{x^{2} + x + 1}}{∣ 1 - x ∣}) - \frac{\sqrt{3}}{3} a r c t a n (\frac{1}{\sqrt{3}} (2 x + 1)) + λ

= A (x) + λ \Rightarrow λ = {l i m}_{\to 0} S (x) - A (x) = - A (0)

= \frac{\sqrt{3}}{3} a r c t a n (\frac{1}{\sqrt{3}}) = \frac{\sqrt{3}}{3} \frac{π}{6} = \frac{π \sqrt{3}}{18} a n d S (x) = \sum_{n = 0}^{\infty} \frac{x^{3 n + 2}}{3 n + 2}

w e h a v e \sum_{n = 0}^{\infty} \frac{1}{(3 n + 2) 3^{n}} = \sum_{n = 0}^{\infty} \frac{{(\frac{1}{3})}^{n}}{3 n + 2}

= {(3_{\sqrt{3}})}^{2} \sum_{n = 0}^{\infty} \frac{{(\frac{1}{3_{\sqrt{3}}})}^{3 n + 2}}{3 n + 2} = {(3_{\sqrt{3}})}^{2} S ({(3_{\sqrt{3}})}^{- 1}) t h e s u m S i s

k n o w n .