# Why $1− \frac12 + \frac13 − \frac14 + \frac15 - \cdots$ is not zero but $\ln2$ [duplicate]

What's wrong with this:

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...$$ $$=(1+\frac{1}{3}+\frac{1}{5}+...)-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)$$ $$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...) - 2 (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)$$ $$=(1+\frac{1}{2}+\frac{1}{3}+...) - (1+\frac{1}{2}+\frac{1}{3}+...)$$ $$= 0$$

The mistake in this argument is in the second equality, where it is claimed that $$(1+\frac{1}{3}+\frac{1}{5}+\cdots)-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots) =(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots) - 2 (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots).$$ This is not true because the two series on the left-hand side do not converge absolutely. In fact, both series diverge, so their difference is undefined. Therefore, the subsequent steps in the argument are also invalid. The series $$\displaystyle1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is known as the alternating harmonic series, and it is conditionally convergent, which means that it converges but not absolutely. This means that the order in which the terms are added can affect the sum, as demonstrated by the incorrect argument above.
Indeed, the Taylor series expansion of $$\ln(1+x)$$ is: $$\ln(1+x) = x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4 + \cdots$$ If we set $$x=1$$, we get: $$\displaystyle\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$