$\ln2\approx.693$, according to my calculator. It can be written as the infinite sum $$1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19-\frac1{10}\dots$$

Rearranging this infinite sum by odds and evens gives:


This is the same as:


Combining the first two parentheses:

$$\left(1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\frac19+\frac1{10}\dots\right) -2\left(\frac12+\frac14+\frac16+\frac18+\frac1{10}\dots\right)$$

Distributing the 2:


But we started out with the expansion of $\ln2\approx.693$. So how did we go from that to $0$? Where did I make my mistake?

  • 12
    $\begingroup$ You can't always rearrange conditionally convergent series like that $\endgroup$ – Joe Aug 2 '18 at 21:53
  • 9
    $\begingroup$ The original infinite sum is conditionally convergent. You can rearrange it to get any real value you choose or to diverge to $\pm \infty$. $\endgroup$ – Mark Bennet Aug 2 '18 at 21:53
  • 6
    $\begingroup$ See Riemann Rearrangement Theorem $\endgroup$ – Sri-Amirthan Theivendran Aug 2 '18 at 22:06
  • 1
    $\begingroup$ You can rearrange a conditionally but not absolutely convergent sequence to get any number you want... $\endgroup$ – copper.hat Aug 2 '18 at 22:39
  • 1
    $\begingroup$ @MichaelHardy All my calculator told me is that if I type in $\ln2$ it equals about .693. The convergent series is just a well-known series that converges to this same figure. $\endgroup$ – DonielF Aug 2 '18 at 22:44

$$ 1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19-\frac1{10} + \cdots $$ Note that \begin{align} 1 + \frac 1 3 + \frac 1 5 + \frac 1 7 + \frac 1 9 + \frac 1 {11} \cdots & =+\infty \\[10pt] \text{and } -\frac 1 2 - \frac 1 4 - \frac 1 6 - \frac 1 8 - \frac 1 {10} - \cdots & = -\infty \end{align} When the positive and negative parts of a convergent series both diverge to infinity, and only then, the value of the sum can be altered by rearranging the terms, i.e. adding in a different order.

That can be seen as follows: Suppose, for example, that I want to make the series converge to $3.$ Since the positive terms add up to $+\infty,$ at some point in adding them you'll get a number bigger than $3.$ Then add the first negative term. Since it's bigger than the last positive term you added, you'll a number less than $3.$ Then keep adding positive terms until it's more than $3.$ Then add the next negative term. And so on.

  • 1
    $\begingroup$ What if all terms are $\pm 4$? Can one rearrange to get 3? I think one needs some more conditions on the positive/negative terms... $\endgroup$ – coffeemath Aug 2 '18 at 23:22
  • 2
    $\begingroup$ @coffeemath: The condition you need is that the terms converge to zero. $\endgroup$ – user14972 Aug 3 '18 at 2:25
  • $\begingroup$ This is Rieeman rearrangement theorem. Since the sum converges only conditionally, for any $\alpha \in \mathbb{R}$, we can rearrange our sum to obtain $\alpha$. $\endgroup$ – user518441 Aug 3 '18 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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