Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question

marked as duplicate by J. M., Ross Millikan, Did, Zev Chonoles Jul 12 '12 at 17:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Possible duplicate of this question. Also related. – J. M. Jul 12 '12 at 4:15
The second equality with ln(2) was broken, the "however" should be After the infinite sum. – The Substitute Jul 12 '12 at 4:31
up vote 5 down vote accepted

Nice argument! You have produced a special case of the Riemann Rearrangement Theorem.

If we have a conditionally convergent series, the terms can be rearranged so that the sum is anything we like, or so that the sum does not exist. You have rearranged a conditionally convergent series for $\log 2$ so that it has sum $\frac{1}{2}\log 2$.

Remark: Results that are clear for finite sums do not necessarily extend to infinite "sums." And continuous functions do not necessarily behave like the smooth curves of our imagination. That is part of what makes analysis necessary.

share|cite|improve this answer
Thank you, I will look into that Theorem. – The Substitute Jul 12 '12 at 4:39

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.

share|cite|improve this answer

You have rediscovered the difference between absolute convergence and conditional convergence, and the fact that rearrangement of conditionally convergent series can alter the sum.

If the sum of the positive terms and the sum of the negative terms are both finite, then the series converges absolutely and rearrangement won't change the sum.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.