Summation of series and Taylor series are giving different results $A = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots$ 
$B = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots$ 
$A + B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$ 
$2A = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots$ 
$2A = A + B$
$A - B = 0$
$0 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$  
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 ?

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. 

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 $\lim_{N\to\infty}(A_N-B_N)=\log(1/2)$).
