The question is to evaluate $$\sum_{n=1}^\infty \frac1{n(2n-1)}$$
I have done this:
$$\sum_1^\infty\frac1{n(2n-1)}=\sum_{n=1}^\infty\frac2{2n-1}-\frac1n=\frac21-\frac11+\frac23-\frac12+\cdots\\=1-\frac12+\frac13-\frac14+\cdots=\log 2$$ But Wolfram alpha says the initial sum is $2\log 2$. I think the wrong step is when I started a new line above - the sum needs to be re-ordered to get it into the form I have got. But the series is not absolutely convergent (I think), so this shouldn't be allowed. Am I right? How can I correct this argument? If it is not possible to correct is, can someone provide an alternative method I could use instead?