beero1000
Veteran Member
I covered the Riemann rearrangement theorem in class today, and figured that the boards don't have enough 'infinity' threads.
The alternating harmonic series \(\sum_{n=1}^\infty \left(-1\right)^{n+1}\left(\frac{1}{n}\right)\) is well known to converge to \(\ln 2\).
On the other hand, by rearranging the terms of \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots\) we get \(\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = (1 - \frac{1}{2}) - \frac{1}{4} + (\frac{1}{3} - \frac{1}{6}) - \frac{1}{8} + (\frac{1}{5} - \frac{1}{10}) + \dots = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \dots = \frac{1}{2}(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots) = \frac{1}{2} \ln 2\). In other words, \(1 = 2\).
Of course, that isn't quite kosher, but what about the series \(10 - 1 + 10 - 1 + 10 - 1 + \dots\)?
The alternating harmonic series \(\sum_{n=1}^\infty \left(-1\right)^{n+1}\left(\frac{1}{n}\right)\) is well known to converge to \(\ln 2\).
On the other hand, by rearranging the terms of \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots\) we get \(\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = (1 - \frac{1}{2}) - \frac{1}{4} + (\frac{1}{3} - \frac{1}{6}) - \frac{1}{8} + (\frac{1}{5} - \frac{1}{10}) + \dots = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \dots = \frac{1}{2}(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots) = \frac{1}{2} \ln 2\). In other words, \(1 = 2\).
Of course, that isn't quite kosher, but what about the series \(10 - 1 + 10 - 1 + 10 - 1 + \dots\)?