find-the-value-of-n-0-1-3-n-n-1-n-2-

Question Number 31749 by abdo imad last updated on 13/Mar/18

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

Commented by abdo imad last updated on 14/Mar/18

l e t p u t S (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{(n + 1) (n + 2)} = \sum_{n = 0}^{\infty} a_{n} x^{n}

l e t f i n d t h e r a d i u s o f t h i s s e r i e w e h a v e

\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1) (n + 2)}{(n + 2) (n + 3 {} \Rightarrow {l i m}_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 1 s o f o r ∣ x ∣< 1

t h e s e r i e i s c o n v e r g e n t w e h a v e

S (x) = \sum_{n = 0}^{\infty} (\frac{1}{n + 1} - \frac{1}{n + 2}) x^{n} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n + 1} - \sum_{n = 0}^{\infty} \frac{x^{n}}{n + 2}

= \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} - \sum_{n = 2}^{\infty} \frac{x^{n - 2}}{n}

= \frac{1}{x} \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - \frac{1}{x^{2}} (\sum_{n = 1}^{\infty} \frac{x^{n}}{n} - x)

= \frac{1}{x} + (\frac{1}{x} - \frac{1}{x^{2}}) \sum_{n = 1}^{\infty} \frac{x^{n}}{n} l e t p u t f (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

f^{^{'}} (x) = \sum_{n = 1}^{\infty} x^{n - 1} = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow f (x) = - l n ∣ 1 - x ∣ + λ

λ = f (0) = 0 \Rightarrow \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \Rightarrow S (x) = \frac{1}{x} - (\frac{1}{x} - \frac{1}{x^{2}}) l n ∣ 1 - x ∣

\sum_{n = 0}^{\infty} \frac{1}{3^{n} (n + 1) (n + 2)} = S (\frac{1}{3})

= 3 - (3 - 9) l n ∣ \frac{2}{3} ∣= 3 + 6 (l n (2) - l n (3)) .

Commented by rahul 19 last updated on 14/Mar/18

c a n u p r o v i d e s o l . f o r t h i s o n e ?

I t r i e d a s :

\underset{n = 0}{\sum^{\infty}} (\frac{1}{n + 1} - \frac{1}{n + 2}) \frac{1}{3^{n}} b u t c o u l d n o t f i n d

a n y p a t t e r n f u r t h e r .