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# How do you find the sum of ln ( 1 + 1 / (2^(2^n)) ) from n=0 to infinity? Thanks!?

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#### Explanation

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#### Explanation:

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1
Feb 9, 2018

$\setminus$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus {\lim}_{n \rightarrow \setminus \infty} {\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right) \setminus = \setminus \ln \left(2\right) .$

#### Explanation:

$\setminus$

$\text{ Let's start with the observation:}$

$\left(1 - \frac{1}{{2}^{{2}^{n}}}\right) \left(1 + \frac{1}{{2}^{{2}^{n}}}\right) \setminus = \setminus \left(1 - {\left(\frac{1}{{2}^{{2}^{n}}}\right)}^{2}\right)$

$\setminus q \quad \setminus q \quad = \setminus 1 - \frac{1}{{2}^{2 \setminus \cdot {2}^{n}}} \setminus = \setminus 1 - \frac{1}{{2}^{{2}^{n + 1}}}$

$\text{ Thus:}$

$\left(1 - \frac{1}{{2}^{{2}^{n}}}\right) \left(1 + \frac{1}{{2}^{{2}^{n}}}\right) \setminus = \setminus 1 - \frac{1}{{2}^{{2}^{n + 1}}} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(1\right)$

$\text{ So, iterating this:}$

$\left(1 - \frac{1}{{2}^{{2}^{0}}}\right) \left(1 + \frac{1}{{2}^{{2}^{0}}}\right) \left(1 + \frac{1}{{2}^{{2}^{1}}}\right) \left(1 + \frac{1}{{2}^{{2}^{2}}}\right) \setminus \cdots \left(1 + \frac{1}{{2}^{{2}^{n}}}\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus 1 - \frac{1}{{2}^{{2}^{n + 1}}} .$

$\setminus$

$\text{I.e.:} \setminus q \quad \setminus q \quad \left(1 - \frac{1}{{2}^{{2}^{0}}}\right) {\prod}_{k = 0}^{n} \left(1 + \frac{1}{{2}^{{2}^{k}}}\right) \setminus = \setminus 1 - \frac{1}{{2}^{{2}^{n + 1}}} . \setminus q \quad \setminus q \quad \setminus q \quad \left(2\right)$

$\text{Taking logarithm of both sides of this equation:}$

$\setminus q \quad \setminus q \quad \ln \left[\left(1 - \frac{1}{{2}^{{2}^{0}}}\right) {\prod}_{k = 0}^{n} \left(1 + \frac{1}{{2}^{{2}^{k}}}\right)\right] \setminus = \setminus \ln \left[1 - \frac{1}{{2}^{{2}^{n + 1}}}\right] .$

$\therefore \setminus \ln \left[\left(1 - \frac{1}{{2}^{{2}^{0}}}\right)\right] + \left[{\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right)\right] \setminus = \setminus \ln \left[1 - \frac{1}{{2}^{{2}^{n + 1}}}\right] .$

$\therefore \setminus {\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right) \setminus = \setminus \ln \left[1 - \frac{1}{{2}^{{2}^{n + 1}}}\right] - \ln \left[1 - \frac{1}{{2}^{{2}^{0}}}\right] .$

$\therefore \setminus {\lim}_{n \rightarrow \setminus \infty} {\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus {\lim}_{n \rightarrow \setminus \infty} \left[\ln \left[1 - \frac{1}{{2}^{{2}^{n + 1}}}\right] - \ln \left[1 - \frac{1}{{2}^{{2}^{0}}}\right]\right] .$

$\therefore \setminus \quad \setminus \quad \setminus {\lim}_{n \rightarrow \setminus \infty} {\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right) \setminus = \setminus \ln \left[1 - 0\right] - \ln \left[1 - \frac{1}{2}\right] .$

$\therefore \setminus \quad \setminus \quad \setminus {\lim}_{n \rightarrow \setminus \infty} {\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right) \setminus = \setminus \ln \left(1\right) - \ln \left(\frac{1}{2}\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus 0 - \left(- \ln \left(2\right)\right) \setminus = \setminus \ln \left(2\right) .$

$\setminus$

$\text{So now we have determined the desired limit.}$

$\text{Statement of result:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus {\lim}_{n \rightarrow \setminus \infty} {\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right) \setminus = \setminus \ln \left(2\right) . \setminus q \quad \setminus \quad \square$

$\setminus$

$\setminus$

$\text{Remark:}$

$\text{In the above, we also established a perhaps more subtle result --}$

$\text{we have found closed a formula for the finite version of this sum:}$

$\therefore \setminus {\sum}_{k = 0}^{n} \ln \left(1 + \frac{1}{{2}^{{2}^{k}}}\right) \setminus = \setminus \ln \left[1 - \frac{1}{{2}^{{2}^{n + 1}}}\right] - \ln \left[1 - \frac{1}{{2}^{{2}^{0}}}\right] .$

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