Summation of series and Taylor series are giving different results

Question

$A = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots$
$B = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots$
$A + B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$
$2A = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$
$2A = A + B$
$A - B = 0$
$0 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$

Using Taylor expansion for $\ln(1+x)$
$\ln(2) = 1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$

Where did I go wrong ?

Answer 1

What you did wrong was to assume that you can deal with divergent series as if they were convergent. If you say that

$$ A=\frac12+\frac14+\frac16+\cdots $$

then $A$ is a name of the series, but it is not a number (since the series diverges).

Therefore $$ 1+\frac12+\frac13+\frac14+\cdots $$ is just another (divergent) series and it is an unfortunate option to call it $2A$.

Answer 2

As others have already pointed out, the series $\sum_{n=1}^\infty \frac1n$ and $\sum_{n=1}^\infty \frac1{2n-1}$ diverge and manipulating divergent series is not legitimate.

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Another way to analyze this is to look at the partial sums $A_N=\sum_{n=1}^{N} \frac{1}{2n}$ and $B_N=\sum_{n=1}^N \frac{1}{2n-1}$.

While $A_N+B_N =\sum_{n=1}^{2N} \frac1n$, we have $A_N-B_N=-\sum_{n=N+1}^{2N}\frac1n \ne 0$ (In fact, we have $\lim_{N\to\infty}(A_N-B_N)=\log(1/2)$).