Conditionally Convergent

[beero1000]

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 \(10 - 1 + 10 - 1 + 10 - 1 + \dots\)?

 

[Kharakov]

It appears like the order of the series matters when dealing with infinite series- like you end up further and further into series debt:

\(+ \left( \frac{1}{7} - \frac{1}{14}\right) +\left( \frac{1}{9}-\frac{1}{18} \right) +\left( \frac{1}{11}-\frac{1}{22} \right)...\)

Except for this:

\(\sum_{n=1}^\infty \left(\frac{1}{n}\right)=1 + \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}...\)

\(\frac{1}{2}\times \sum_{n=1}^\infty \left(\frac{1}{n}\right)= \frac{1}{2}+ \frac{1}{4}+ \frac{1}{6}...\)

Subtracting 2 * 1/2 the harmonic series from the harmonic series then gives you ln2.

\(1 + \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}... - \frac{1}{2}- \frac{1}{4}- \frac{1}{6}...= 1 + \frac{1}{3}+ \frac{1}{5}+ \frac{1}{7}... \)

\(1 + \frac{1}{3}+ \frac{1}{5}+ \frac{1}{7}... - \frac{1}{2}- \frac{1}{4}- \frac{1}{6}...= 1 - \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}...\)

In other words, infinity-infinity= ln2.... or whatchamacallit. 6? Indeterminate. Looks sort of like another of the cool indeterminate form things.


I used to think there was some justification in one of the Taylor series for log:

\(log \frac{x}{y} = \sum_{n=1}^\infty \left(\frac {x-y}{x}\right)^n\times \frac{1}{n}\)

ln1-ln2 = -1 +1/2 -1/3 ... so ln2 - ln1= 1 -1/2 + 1/3 ....

but I think the series diverges for difference greater than one (ln1 - ln3)....

Anyway, you end up with a grid of zeta series if you stack log n - log (n-1) =...