Find-the-infinite-sum-n-1-1-2n-1-2n-1-2n-3-

Question Number 90787 by Maclaurin Stickker last updated on 26/Apr/20

F i n d t h e i n f i n i t e s u m

\underset{n = 1}{\sum^{\infty}} \frac{1}{(2 n - 1) (2 n + 1) (2 n + 3)}

Commented by mathmax by abdo last updated on 26/Apr/20

l e t d e c o m p o s e F (x) = \frac{1}{(2 x - 1) (2 x + 1) (2 x + 3)}

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

a = (2 x - 1) F (x) ∣_{x = \frac{1}{2}} = \frac{1}{(2) (4)} = \frac{1}{8}

b = (2 x + 1) F (x) ∣_{x = - \frac{1}{2}} = \frac{1}{(- 2) (2)} = - \frac{1}{4}

c = (2 x + 3) F (x) ∣_{x = - \frac{3}{2}} = \frac{1}{(- 4) (- 2)} = \frac{1}{8} \Rightarrow

F (x) = \frac{1}{8 (2 x - 1)} - \frac{1}{4 (2 x + 1)} + \frac{1}{8 (2 x + 3)} l e t

S_{n} = \sum_{k = 1}^{n} \frac{1}{(2 k - 1) (2 k + 1) (2 k + 3)} \Rightarrow S_{n} = Σ F (k)

= \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{4} \sum_{k = 1}^{n} \frac{1}{2 k + 1} + \frac{1}{8} \sum_{k = 1}^{n} \frac{1}{2 k + 3} w e h a v e

\sum_{k = 1}^{n} \frac{1}{2 k - 1} = 1 + \frac{1}{3} + \frac{1}{5} + \dots . + \frac{1}{2 n - 1}

= 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2 n - 1} + \frac{1}{2 n} - \frac{1}{2} - \frac{1}{4} - \dots . - \frac{1}{2 n}

= H_{2 n} - \frac{1}{2} H_{n}

\sum_{k = 1}^{n} \frac{1}{2 k + 1} =_{k = j - 1} \sum_{j = 2}^{n + 1} \frac{1}{2 j - 1} = \sum_{j = 1}^{n + 1} \frac{1}{2 j - 1} - 1

= H_{2 n + 2} - \frac{1}{2} H_{n + 1} - 1

\sum_{k = 1}^{n} \frac{1}{2 k + 3} =_{k = j - 2} \sum_{j = 3}^{n + 2} \frac{1}{2 j - 1} = \sum_{j = 1}^{n + 2} \frac{1}{2 j - 1} - 1 - \frac{1}{3}

= H_{2 n + 4} - \frac{1}{2} H_{n + 2} - \frac{4}{3} \Rightarrow

8 S_{n} = H_{2 n} - \frac{1}{2} H_{n} - 2 (H_{2 n + 2} - \frac{1}{2} H_{n + 1} - 1) + H_{2 n + 4} - \frac{1}{2} H_{n + 2} - \frac{4}{3}

= H_{2 n} - \frac{1}{2} H_{n} - 2 H_{2 n + 2} + H_{n + 1} + 2 + H_{2 n + 4} - \frac{1}{2} H_{n + 2} - \frac{4}{3}

\sim l n (2 n) + γ - \frac{1}{2} l n (n) - \frac{γ}{2} - 2 l n (2 n + 2) - 2 γ + l n (n + 1) + γ

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

8 S_{n} \sim l n (2) + l n (n) - \frac{1}{2} l n (n) - 2 l n (2) - 2 l n (n) + l n (n)

+ l n (2) + l n (n) - \frac{1}{2} l n (n) + \frac{2}{3} \Rightarrow 8 S_{n} \sim \frac{2}{3} \Rightarrow S_{n} \sim \frac{1}{12}

\Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{12}

Commented by Maclaurin Stickker last updated on 26/Apr/20

T h a n k y o u, I a p p r e c i a t e y o u r w o r k .

Commented by abdomathmax last updated on 27/Apr/20

y o u a r e w e l c o m e

Answered by ajfour last updated on 26/Apr/20

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

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

= \frac{1}{8} Σ \frac{1}{2 n - 1} - \frac{1}{8} Σ \frac{1}{2 n + 1} - \frac{1}{8} Σ \frac{1}{2 n + 1} + \frac{1}{8} Σ \frac{1}{2 n + 3}

8 S = (\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots) - 2 (\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots) + (\frac{1}{5} + \frac{1}{7} + \dots)

= 1 - \frac{1}{3} = \frac{2}{3}

\Rightarrow S = \frac{1}{12} .

(a n y c h a n c e t h i s m a y b e c o r r e c t ?)

Commented by mathmax by abdo last updated on 26/Apr/20

c o r r e c t s i r a j f o u r

Commented by Maclaurin Stickker last updated on 26/Apr/20

P e r f e c t!