mathematical-analysis-prove-that-n-0-1-1-2-2-n-2-

Question Number 141132 by mnjuly1970 last updated on 16/May/21

\dots m a t h e m a t i c a l \dots a n a l y s i s \dots

p r o v e t h a t :

\underset{n = 0}{\prod^{\infty}} (1 + \frac{1}{2^{2^{n}}}) \overset{?}{=} 2

\dots \dots

Answered by Dwaipayan Shikari last updated on 16/May/21

\underset{n = 0}{\prod^{\infty}} (1 + \frac{1}{x^{2^{n}}}) = \underset{n = 0}{\prod^{\infty}} \frac{1 - \frac{1}{x^{2^{n + 1}}}}{1 - \frac{1}{x^{2^{n}}}} = \frac{1 - \frac{1}{x^{2}}}{1 - \frac{1}{x}} . \frac{1 - \frac{1}{x^{4}}}{1 - \frac{1}{x^{2}}} . \frac{1 - \frac{1}{x^{8}}}{1 - \frac{1}{x^{4}}} . \frac{1 - \frac{1}{x^{16}}}{1 - \frac{1}{x^{8}}} \dots \frac{1 - \frac{1}{x^{2^{N + 1}}}}{1 - \frac{1}{x^{2^{N}}}}

= \lim_{N \to \infty} \frac{x}{x - 1} . (1 - \frac{1}{x^{2^{N + 1}}}) = \frac{x}{x - 1}

x = 2

Commented by mnjuly1970 last updated on 16/May/21

v e r y n i c e . .