# "Proof" that $\log2=0$ using the expansion of $\log(1+x)$

$$\log(1+x)=x-\frac{x^2}{2} +\frac{x^3}{3}-\frac{x^4}{4}\cdots$$ Putting $x=1$ $$\log 2=1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18\cdots$$ $$=\left\{\left(1+\frac13+\frac15+\frac17 \cdots\right)-\left(\frac12+\frac14+\frac16 +\frac18 \cdots\right)\right\}$$ $$=\left\{\left(1+\frac13+\frac15+\frac17\cdots\right)+\left(\frac12+\frac14+\frac16 +\frac18\cdots \right)\right\}-2\left(\frac12+\frac14+\frac16 +\frac18 \cdots \right)$$ $$=\left(1+\frac12+\frac13+\frac14+\frac15\cdots\right)-\left(1+\frac12+\frac13+\frac14+\frac15\cdots\right)$$ $$\log 2=0$$ $$2=e^0$$ $$2=1$$ Same result is obtained when I multiplied the $\log2$expansion by $2$ .

• The expansion of $\log(1+x)$ is valid for $|x|<1$ Aug 22, 2016 at 7:09
• @ClaudeLeibovici it is valid for $x=1$ though. Aug 22, 2016 at 7:13
• To put it more formally, $\sum_{k = 0}^{+\infty} (\frac{1}{2k-1}- \frac{1}{2k})$. Since $\sum_{k=0}^{+\infty} \frac{1}{2k-1}$ and $\sum_{k=0}^{+\infty} \frac{1}{2k}$, so the addition is not linear.(i.e. it is invalid to write as$\sum_{k=0}^{+\infty} \frac{1}{2k-1}-\sum_{k=0}^{+\infty} \frac{1}{2k}$ )
– Zau
Aug 22, 2016 at 7:21
• You can change the order of summation of a series and get the same result only if the series converges absolutely. This is not the case here so you get whatever you want (litereraly) by Riemann's THeorem. Aug 22, 2016 at 7:29

Well, there's one mistake along the way that I have spotted: $$\cdots=\left\{\left(1+\frac13+\frac15+\frac17\color{red}{+\cdots}\right)-\left(\frac12+\frac14+\frac16 +\frac18\color{red}{+\cdots} \right)\right\}$$

In both of the smaller parenthesis, you have a variation of the harmonic series, and neither $$1+\frac13+\frac15+\frac17\color{red}{+\cdots}$$ nor $$\left(\frac12+\frac14+\frac16 +\frac18\color{red}{+\cdots} \right)$$ converges. You cannot split an infinite sum to two inifinite sums if either of the three infinite sums does not converge.

You may look in Riemann series theorem The theorem of Riemann says that any conditionally convergent series (yours being the 'classic' example) may be rearranged to converge to any given number (including $\pm \infty$).

If  both series $$\sum_{n=1}^{\infty} a_{2n-1} \qquad\text{and}\qquad \sum_{n=1}^{\infty} a_{2n}$$ are convergent, then also $$\sum_{n=1}^{\infty}(-1)^{n+1}a_n$$ is convergent and $$\sum_{n=1}^{\infty}(-1)^{n+1}a_n= \sum_{n=1}^{\infty} a_{2n-1} - \sum_{n=1}^{\infty} a_{2n}$$ (the proof is not difficult, change sign to the even numbered terms in order to simplify it and get rid of the $(-1)^{n+1}$). There's no requirement for absolute convergence here.

Unfortunately, in your case you cannot apply this result, because neither the odd-numbered term series nor the even-numbered term series converge.

You assumed that

$\left(\frac12+\frac14+\frac16 +\frac18+\cdots \right)=2 \left(\frac12+\frac14+\frac16 +\frac18+\cdots \right)- \left(\frac12+\frac14+\frac16 +\frac18+\cdots \right)$

Which means

$\infty = 2 \infty - \infty$

You are manipulating infinity and this is the mistake.