MHB Evaluate the product ∏(1+10^(-2^n))

[lfdahl]

Evaluate:

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

 

[kaliprasad]

Evaluate:

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

Spoiler

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]

Spoiler

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