### Instruction: Find
\[\prod_{k = 0}^\infty \left( 1 + \frac{1}{14^{2^k}} \right).\].### Response: I see an infinite product of terms of the form \(\left( 1 + \frac{1}{14^{2^k}} \right)\).
This reminds me of a formula for the sum of a geometric series: \(\sum_{k = 0}^\infty a r^k = \frac{a}{1 - r}\) for any constant \(a\) and any \(r\) with \(\lvert r \rvert < 1\).
Maybe I can rewrite the product as a series by taking logarithms.
If I let \(y = \prod_{k = 0}^\infty \left( 1 + \frac{1}{14^{2^k}} \right)\), then \(\ln y = \sum_{k = 0}^\infty \ln \left( 1 + \frac{1}{14^{2^k}} \right)\).
Now I need to find a way to simplify \(\ln \left( 1 + \frac{1}{14^{2^k}} \right)\).
I know that \(\ln (1 + x) \approx x\) for small \(x\), so maybe I can use that as an approximation.
If I do that, I get \(\ln y \approx \sum_{k = 0}^\infty \frac{1}{14^{2^k}}\).
This is a geometric series with \(a = \frac{1}{14}\) and \(r = \frac{1}{14^2}\).
Using the formula for the sum of a geometric series, I get \(\ln y \approx \frac{\frac{1}{14}}{1 - \frac{1}{14^2}} = \frac{1}{14 - \frac{1}{14}} = \frac{1}{\frac{195}{14}} = \frac{14}{195}\).
Therefore, \(y \approx e^{\frac{14}{195}}\).
This is the approximate value of the product.