Prove-that-2-1-4-4-1-8-8-1-16-16-1-32-2-

Question Number 95668 by I want to learn more last updated on 26/May/20

Prove that : 2^{1 / 4} . 4^{1 / 8} . 8^{1 / 16} . 16^{1 / 32} . \dots \infty = 2

Commented by Tony Lin last updated on 26/May/20

2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times 8^{\frac{1}{16}} \times 16^{\frac{1}{32}} \times \cdot \cdot \cdot

= 2^{\frac{1}{4}} \times {(2^{2})}^{\frac{1}{8}} \times {(2^{3})}^{\frac{1}{16}} \times {(2^{4})}^{\frac{1}{32}} \times \cdot \cdot \cdot = 2^{1}

s o w h a t w e n e e d t o p r o v e i s

\frac{1}{2^{2}} + \frac{2}{2^{3}} + \frac{3}{2^{4}} + \frac{4}{2^{5}} + \cdot \cdot \cdot = 1

\Rightarrow \underset{k = 1}{\sum^{\infty}} \frac{k}{2^{k + 1}} = 1

l e t f (x) = \underset{k = 1}{\sum^{\infty}} k \frac{x^{k - 1}}{2^{k + 1}}

l e t F (x) = \int_{0}^{x} f (t) d t = \underset{k = 1}{\sum^{\infty}} \frac{x^{k}}{2^{k + 1}} + c

= \frac{1}{2} [{(\frac{x}{2})}^{1} + {(\frac{x}{2})}^{2} + {(\frac{x}{2})}^{3} + \cdot \cdot \cdot] + c

= \frac{1}{2} \times \frac{\frac{x}{2}}{1 - \frac{x}{2}} + c

= \frac{x}{4 - 2 x} + c

w h e n x = 0, F (x) = 0

∴ c = 0

\Rightarrow F (x) = \frac{x}{4 - 2 x}

f (x) = F^{'} (x) = \frac{4}{{(4 - 2 x)}^{2}}

p l u g x = 1

\Rightarrow \underset{k = 1}{\sum^{\infty}} \frac{k}{2^{k + 1}} = \frac{4}{{(4 - 2)}^{2}} = 1

h e n c e p r o v e d

Commented by I want to learn more last updated on 06/Jun/20

Thanks sir

Answered by Rio Michael last updated on 26/May/20

my try please check .

2^{1 / 4} . 4^{1 / 8} . 8^{1 / 16} . 16^{1 / 32} \dots \infty = \underset{n = 1}{\prod^{\infty}} 2^{\frac{n}{2^{n + 1}}}

now let x = \underset{n = 1}{\prod^{\infty}} 2^{\frac{n}{2^{n + 1}}}

\Leftrightarrow \ln x = \underset{n = 1}{\sum^{\infty}} \frac{n}{2^{n + 1}} \ln 2 \dots \dots \dots . (i)

\ln x = \ln 2 \underset{n = 1}{\sum^{\infty}} \frac{n}{2^{n + 1}} = \frac{1}{2^{2}} + \frac{2}{2^{3}} + \frac{3}{2^{4}} + \dots

= \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \dots . is a GP with r = \frac{1}{2} \Rightarrow S_{\infty} = \frac{1}{16}

\Rightarrow \underset{n = 1}{\sum^{\infty}} 2^{\frac{n}{n + 1}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

= \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{4} + \dots) = \frac{1}{2} \frac{1}{1 - \frac{1}{2}} = 1 \dots \dots \dots (i i)

eqn (i) + eqn (i i) \Rightarrow \ln x = \ln 2 \times 1 = \ln 2 \Leftrightarrow x = 2

∴ \underset{n = 1}{\prod^{\infty}} 2^{\frac{n}{2^{n + 1}}} = 2^{1 / 4} . 4^{1 / 8} . 8^{1 / 16} . 16^{1 / 32} \dots . \infty = 2 .

Commented by I want to learn more last updated on 06/Jun/20

Thanks sir