1-find-the-radius-of-convergence-for-n-1-x-n-n-n-1-n-2-and-calculate-its-sum-2-find-the-value-of-n-1-1-n-n-2-n-n-1-n-2-

Question Number 33710 by math khazana by abdo last updated on 22/Apr/18

1) f i n d t h e r a d i u s o f c o n v e r g e n c e ? f o r

\sum_{n = 1}^{\infty} \frac{x^{n}}{n (n + 1) (n + 2)} a n d c a l c u l a t e i t s s u m

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

Commented by prof Abdo imad last updated on 27/Apr/18

l e t p u t u_{n} = \frac{1}{n (n + 1) (n + 2)} w e h a v e

\frac{u_{n + 1}}{u_{n}} = \frac{n (n + 1) (n + 2)}{(n + 1) (n + 2) (n + 3)} \Rightarrow {l i m}_{n \to + \infty} \frac{u_{n + 1}}{u_{n}} = 1

\Rightarrow R = 1 l e t d e c o m p o s e F (x) = \frac{1}{x (x + 1) (x + 2)}

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

a = {l i m}_{x \to 0} x F (x) = \frac{1}{2}

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

c = {l i m}_{x \to - 2} (x + 2) F (x) = \frac{1}{(- 2) (- 1)} = \frac{1}{2}

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

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

= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - \sum_{n = 1}^{\infty} \frac{x^{n}}{n + 1} + \frac{1}{2} \sum_{n = 1}^{\infty} \frac{x^{n}}{n + 2} b u t

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

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

= \frac{1}{x} (- l n ∣ 1 - x ∣ - x)

\sum_{n = 1}^{\infty} \frac{x^{n}}{n + 2} = \sum_{n = 3}^{+ \infty} \frac{x^{n - 2}}{n} = \frac{1}{x^{2}} \sum_{n = 3}^{+ \infty} \frac{x^{n}}{n}

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

= - \frac{l n ∣ 1 - x ∣}{x^{2}} - \frac{1}{x} - \frac{1}{2} \Rightarrow

S (x) = - \frac{1}{2} l n ∣ 1 - x ∣ - \frac{1}{x} (- l n ∣ 1 - x ∣ - x)

- \frac{l n ∣ 1 - x ∣}{2 x^{2}} - \frac{1}{2 x} - \frac{1}{4}

S (x) = (- \frac{1}{2} + \frac{1}{x} - \frac{1}{2 x^{2}}) l n ∣ 1 - x ∣ + 1 - \frac{1}{2 x} - \frac{1}{4}

S (x) = (- \frac{1}{2} + \frac{1}{x} - \frac{1}{2 x^{2}}) l n ∣ 1 - x ∣ - \frac{1}{2 x} + \frac{3}{4} .

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

= (- \frac{1}{2} - 2 - 2) l n (\frac{3}{2}) + 1 + \frac{3}{4}

= (- \frac{1}{2} - 4) l n (\frac{3}{2}) + \frac{7}{4}

= \frac{7}{4} - \frac{9}{2} (l n (3) - l n (2)) .