find-the-sum-of-n-1-1-n-2-n-

Question Number 26401 by abdo imad last updated on 25/Dec/17

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

\sum_{n = 1}^{\propto} \frac{1}{n 2^{n}} .

Commented by abdo imad last updated on 25/Dec/17

l e t p u t S = \sum_{n = 1}^{\propto} \frac{1}{n 2^{n}} S (x) = \sum_{n = 1}^{\propto} \frac{x^{n}}{n} d u e t o u n i f o r m c o n v e r g e n c e

o f t h a t s e r i e / x / < 1 a n d f o r h i d d e r i v a t i v e w e h a v e

S^{,} (x) = \sum_{n ⩾ 1} x^{n - 1} = \frac{1}{1 - x} \Rightarrow S (x) = - l n / 1 - x / + k

k = S (0) = 0 \Rightarrow S (x) = - l n / 1 - x / s o S = S (\frac{1}{2}) = l n (2)

Answered by prakash jain last updated on 25/Dec/17

\ln (1 - x) = - x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \dots

p u t x = \frac{1}{2}

\ln (1 - \frac{1}{2}) = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n \cdot 2^{n}}

\ln 1 - \ln 2 = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n \cdot 2^{n}}

\ln 2 = \underset{n = 1}{\sum^{\infty}} \frac{1}{n \cdot 2^{n}}

a n s : \ln 2