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$\ln2\approx.693$, 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:

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

This is the same as:

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

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:

$$\left(1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\frac19+\frac1{10}\dots\right)-\left(1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\frac19+\frac1{10}\dots\right)=0$$

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?

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    $\begingroup$ You can't always rearrange conditionally convergent series like that $\endgroup$ – Joe Aug 2 '18 at 21:53
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    $\begingroup$ 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
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    $\begingroup$ See Riemann Rearrangement Theorem $\endgroup$ – Sri-Amirthan Theivendran Aug 2 '18 at 22:06
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    $\begingroup$ 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
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    $\begingroup$ @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
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$$ 1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19-\frac1{10} + \cdots $$ 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.

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    $\begingroup$ 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
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    $\begingroup$ @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

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