Find-the-sum-to-infinity-whose-n-th-term-is-n-2-n-1-

Question Number 46860 by 786786AM last updated on 01/Nov/18

Find the sum to infinity whose n^{th} term is \frac{n}{2^{n - 1}} .

Commented by maxmathsup by imad last updated on 01/Nov/18

l e t s (x) = \sum_{n = 1}^{\infty} {n x}^{n - 1} w i t h ∣ x ∣< 1 w e h a v e \sum_{n = 1}^{\infty} \frac{n}{2^{n - 1}} = s (\frac{1}{2}) b u t

\sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow \sum_{n = 1}^{\infty} {n x}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow s (x) = \frac{1}{{(1 - x)}^{2}} \Rightarrow

s (\frac{1}{2}) = \frac{1}{{(1 - \frac{1}{2})}^{2}} = 4 \Rightarrow \sum_{n = 1}^{\infty} {n x}^{n - 1} = 4 .

Commented by maxmathsup by imad last updated on 02/Nov/18

a n t h e r m e t h o d l e t S_{n} (x) = \sum_{k = 1}^{n} k x^{k - 1} w i t h ∣ x ∣< 1 w e h a v e

S_{\infty} (\frac{1}{2}) = \sum_{k = 1}^{\infty} k x^{k - 1} l e t d e t e r m i n e S_{n} (x) w e h a v e

\sum_{k = 0}^{n} x^{k} = \frac{x^{n + 1} - 1}{x - 1} \Rightarrow \sum_{k = 1}^{n} k x^{k - 1} = \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(1 - x)}^{2}} = S_{n} (x) \Rightarrow

S_{n} (\frac{1}{2}) = 4 {\frac{n}{2^{n + 1}} - \frac{n + 1}{2^{n}} + 1} \to 4 (n \to + \infty) \Rightarrow S_{\infty} (\frac{1}{2}) = 4 = \sum_{n = 1}^{\infty} \frac{n}{2^{n - 1}}

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18

S_{n} = \frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + \dots + \frac{n}{2^{n - 1}}

\frac{S_{n}}{2} = + \frac{1}{2^{1}} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \dots + \frac{n - 1}{2^{n - 1}} + \frac{n}{2^{n}}

S_{n} - \frac{S_{n}}{2} = (\frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n - 1}}) - \frac{n}{2^{n}}

\frac{S_{n}}{2} = \frac{\frac{1}{2^{0}} (1 - \frac{1}{2^{n}})}{1 - \frac{1}{2}} - \frac{n}{2^{n}}

S_{n} = 2 \times \frac{(1 - \frac{1}{2^{n}})}{\frac{1}{2}} - \frac{n}{2^{n}}

w h e n n \to \infty S_{\infty} = 4