Birthday problem without electronic aids in a nuclear winter

This isn't great but should be possible by hand.

Since

$$\ln(x)= \sum_{n=1}^\infty \frac{1}{n}\left(\frac{x-1}{x}\right)^n$$

we have

$$\ln(2) = \sum_{n=1}^\infty \frac{1}{n \cdot 2^n} = \sum_{n=1}^{15}\frac{1}{n \cdot 2^n} + \sum_{n=16}^\infty \frac{1}{n \cdot 2^n}$$

For the first fifteen term, you can do some tedious arithmetic to get

$$\sum_{n=1}^{15} \frac{1}{n \cdot 2^n} = \frac{31972079}{46126080} = 0.6931453745\overline{906870}$$

For the tail, you can approximate as follows:

$$\sum_{n=16}^\infty \frac{1}{n \cdot 2^n} < \frac{1}{16}\sum_{n=16}^\infty \frac{1}{2^n} = \frac{1}{524288} = 0.0000019073486328125$$

Adding these together you get

$$\ln(2) < 0.6931472819393199031\overline{870906} < 0.6\overline{93150684} = \frac{235}{365}$$

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Well, that's still pretty bad, but I've already typed it, so might as well post.