find the value of Σ_(n=0) ^∞ (1/(3^n (n+1)(n+2))) .


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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
$let put S(x)=Σ_(n=0) ^∞ (x^n /((n+1)(n+2)))=Σ_(n=0) ^∞ a_n x^n let find the radius of this serie we have (a_(n+1) /a_n ) =(((n+1)(n+2))/((n+2)(n+3{)) ⇒ lim_(n→∞) (a_(n+1) /a_n ) =1 so for ∣x∣<1 the serie is convergent we have S(x)=Σ_(n=0) ^∞ ((1/(n+1)) −(1/(n+2)))x^n =Σ_(n=0) ^∞ (x^n /(n+1)) −Σ_(n=0) ^∞ (x^n /(n+2)) =Σ_(n=1) ^∞ (x^(n−1) /n) −Σ_(n=2) ^∞ (x^(n−2) /n) =(1/x) Σ_(n=1) ^∞ (x^n /n) −(1/x^2 ) (Σ_(n=1) ^∞ (x^n /n) −x) =(1/x) +((1/x) −(1/x^2 ))Σ_(n=1) ^∞ (x^n /n) let put f(x)=Σ_(n=1) ^∞ (x^n /n) f^′ (x)= Σ_(n=1) ^∞ x^(n−1) =Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒f(x)=−ln∣1−x∣ +λ λ=f(0)=0 ⇒ Σ_(n=1) ^∞ (x^n /n) ⇒ S(x)= (1/x) −((1/x)−(1/x^2 ))ln∣1−x∣ Σ_(n=0) ^∞ (1/(3^n (n+1)(n+2))) =S((1/3)) =3 −(3 −9)ln∣(2/3)∣= 3+6(ln(2)−ln(3)) .$
$l e t p u t S (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{(n + 1) (n + 2)} = \sum_{n = 0}^{\infty} a_{n} x^{n}$ $l e t f i n d t h e r a d i u s o f t h i s s e r i e w e h a v e$ $\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1) (n + 2)}{(n + 2) (n + 3 {} \Rightarrow {l i m}_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 1 s o f o r ∣ x ∣< 1$ $t h e s e r i e i s c o n v e r g e n t w e h a v e$ $S (x) = \sum_{n = 0}^{\infty} (\frac{1}{n + 1} - \frac{1}{n + 2}) x^{n} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n + 1} - \sum_{n = 0}^{\infty} \frac{x^{n}}{n + 2}$ $= \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} - \sum_{n = 2}^{\infty} \frac{x^{n - 2}}{n}$ $= \frac{1}{x} \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - \frac{1}{x^{2}} (\sum_{n = 1}^{\infty} \frac{x^{n}}{n} - x)$ $= \frac{1}{x} + (\frac{1}{x} - \frac{1}{x^{2}}) \sum_{n = 1}^{\infty} \frac{x^{n}}{n} l e t p u t f (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n}$ $f^{^{'}} (x) = \sum_{n = 1}^{\infty} x^{n - 1} = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow f (x) = - l n ∣ 1 - x ∣ + λ$ $λ = f (0) = 0 \Rightarrow \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \Rightarrow S (x) = \frac{1}{x} - (\frac{1}{x} - \frac{1}{x^{2}}) l n ∣ 1 - x ∣$ $\sum_{n = 0}^{\infty} \frac{1}{3^{n} (n + 1) (n + 2)} = S (\frac{1}{3})$ $= 3 - (3 - 9) l n ∣ \frac{2}{3} ∣= 3 + 6 (l n (2) - l n (3)) .$
Commented by rahul 19 last updated on 14/Mar/18

$c a n u p r o v i d e s o l . f o r t h i s o n e ?$ $I t r i e d a s :$ $\underset{n = 0}{\sum^{\infty}} (\frac{1}{n + 1} - \frac{1}{n + 2}) \frac{1}{3^{n}} b u t c o u l d n o t f i n d$ $a n y p a t t e r n f u r t h e r .$
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