A limit unforeseen

We know that

$\lim\limits_{n \to \infty} k^{\frac{1}{n}} = 1$

for any real number $k$ .

Since our problem seeks for a limit similar to that, we just have to show whether the expression inside the exponential power $\frac{1}{n}$ actually converges.

The product

$\displaystyle \prod\limits_{n=1}^{\infty} \Big( 1 - 2^{-n} \Big)$

converges if and only if the sum

$\displaystyle \sum\limits_{n=1}^{\infty} \ln\Big( 1- 2^{-n} \Big)$

converges as well.

Now, note that $\displaystyle\sum\limits_{n=1}^{\infty} 2^{-n}$ converges. Furthermore, we can see that

$\lim\limits_{n \to \infty} \frac{ -\ln( 1 - 2^{-n})}{2^{-n}}$

exists. (See note below)

So, by the limit comparison test, we can now say that $\displaystyle \sum\limits_{n=1}^{\infty} \ln\Big( 1- 2^{-n} \Big)$ converges, making the product

$\displaystyle \prod\limits_{n=1}^{\infty} \Big( 1 - 2^{-n} \Big)$

converge to some value $k$ . Now that we have verified the convergence of the product in the radicand, we can now say that the limit of our original expression exists, and is $1$ .

Note:

We have used $-\ln(1-2^{-n})$ for our series in the numerator of the limit comparison test, as the original sum is negative all throughout. Now, To find $\displaystyle \lim\limits_{n \to \infty} \frac{-\ln(1- 2^{-n})}{2^{-n}}$ , we first let $x = 2^{-n}$ , simplifying our limit to

$\lim\limits_{x \to 0} \frac{-\ln(1-x)}{x}$

which is now solvable by L'hopital's rule, giving us $1$ .

If you look at the form it is something tending towards $1^0$ . Which is not an indeterminant form. Any finite no. When raised to zero gives $1$ as output