find-n-1-1-n-n-1-2-n-1-

Question Number 33124 by prof Abdo imad last updated on 10/Apr/18

f i n d \sum_{n = 1}^{\infty} \frac{1}{n (n + 1) 2^{n - 1}} .

Commented by prof Abdo imad last updated on 15/Apr/18

l e t c o n s i d e r S (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n (n + 1)} w i t h ∣ x ∣< 1

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

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

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

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

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

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

= \frac{1 - x}{x} l n ∣ 1 - x ∣ - 1 f o r x = \frac{1}{2} w e g e t

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

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

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