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Evaluate the product ∏(1+10^(-2^n))

lfdahl

Well-known member
Nov 26, 2013
719
Evaluate:

$$\prod_{n=1}^{\infty}\left(1+10^{-2^n}\right)$$
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
Evaluate:

$$\prod_{n=1}^{\infty}\left(1+10^{-2^n}\right)$$

Llet $x=\prod_{n=1}^{\infty}
(1+10^{-2^n})$
Using $(1-10^{-2^n})(1+10^{-2^n}) = (1+10^{-2^{n+1}})$
We have
$x(1-10^{-2^1})=(1-10^{-2^1})\prod_{n=1}^{\infty}(1+10^{-2^n})$
$=(1-10^{-2^2})\prod_{n=2}^{\infty}(1+10^{-2^n})$
$=\lim{n=\infty}(1+10^{-2^n}) = 1$
or x * .99 = 1 or $x = \frac{1}{.99}=\frac{100}{99}$

 

lfdahl

Well-known member
Nov 26, 2013
719

Llet $x=\prod_{n=1}^{\infty}
(1+10^{-2^n})$
Using $(1-10^{-2^n})(1+10^{-2^n}) = (1+10^{-2^{n+1}})$
We have
$x(1-10^{-2^1})=(1-10^{-2^1})\prod_{n=1}^{\infty}(1+10^{-2^n})$
$=(1-10^{-2^2})\prod_{n=2}^{\infty}(1+10^{-2^n})$
$=\lim{n=\infty}(1+10^{-2^n}) = 1$
or x * .99 = 1 or $x = \frac{1}{.99}=\frac{100}{99}$

Thankyou for a clever solution, kaliprasad ! - and for your participation