A misbehaved Power Series (contradiction with ln(2)) 
Possible Duplicate:
Explain why calculating this series could cause paradox? 

Using the power series expansion $\ln(x+1)=\sum _{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$ we have $\ln(2)=(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+\cdots+(\frac{1}{2k-1}-\frac{1}{2k})+\cdots=\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$. Let's rearrange the terms of this sequence so that two negative terms follow each positive term:
$\ln(2)=(1-\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{6}-\frac{1}{8})+\cdots+(\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})+\cdots=\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})$. However, $\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k})$; therefore, $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})=\frac{1}{2}\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$, but how can the latter inequality hold if $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{2k})$ and $\sum _{k=1}^\infty (\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k})$ are equal to $\ln(2)$?
Feel free to come up with a better title...

The series for $\ln 2$ is a conditionally convergent series.  An absolutely convergent series is one that still converges if you take the absolute value of all the terms.  This one is not.  If you take the absolute value of all the terms, you get the harmonic series, which diverges.  Only absolutely convergent series can be rearranged in order and still sum to the same value.  Rearranging conditionally convergent series can give any value you want, as shown by the Riemann rearrangement theorem.
