Log 2 equals to 0

Calculus Level 2

Consider the harmonic series: x=11x=11+12+13+14+15+ \displaystyle \sum_{x=1}^\infty \frac1x = \frac11 + \frac12 + \frac13 + \frac14 + \frac15 + \ldots

I'm going to rearrange the terms to obtain ln(2)=0\ln(2) = 0 by first grouping the terms in pairs:

x=11x=(11+12)+(13+14)+(15+16)+=x=1(12x1+12x)=x=1(2x+2x12x(2x1))=x=1(2(2x1)+12x(2x1))=x=1[2(2x1)2x(2x1)+12x(2x1)]=x=1[1x+12x(2x1)]x=11x=x=11x+x=112x(2x1)0=x=112x(2x1)0=x=1(12x12x1)0=112+1314+15=ln(2)\begin{aligned} \displaystyle \sum_{x=1}^\infty \frac1x &=& \left( \frac11 + \frac12 \right)+\left( \frac13 + \frac14 \right)+\left( \frac15 + \frac16 \right)+\ldots \\ \displaystyle &=& \sum_{x=1}^\infty \left( \frac1{2x-1} + \frac1{2x} \right) \\ \displaystyle &=& \sum_{x=1}^\infty \left( \frac{2x+2x-1}{2x(2x-1)} \right) \\ \displaystyle &=& \sum_{x=1}^\infty \left( \frac{2(2x-1) + 1}{2x(2x-1)} \right) \\ \displaystyle &=& \sum_{x=1}^\infty \left [ \frac{2(2x-1)}{2x(2x-1)} + \frac1{2x(2x-1)} \right] \\ \displaystyle &=& \sum_{x=1}^\infty \left [ \frac1x + \frac1{2x(2x-1)} \right] \\ \displaystyle \xcancel{\sum_{x=1}^\infty \frac1x} &=& \xcancel{\sum_{x=1}^\infty \frac1x} + \sum_{x=1}^\infty \frac1{2x(2x-1)} \\ \displaystyle 0 &=& \sum_{x=1}^\infty \frac1{2x(2x-1)} \\ \displaystyle 0 &=& \sum_{x=1}^\infty \left ( \frac1{2x} - \frac1{2x-1} \right ) \\ \displaystyle 0 &=& 1 - \frac12 + \frac13 - \frac14 + \frac15 - \ldots = \ln(2) \\ \end{aligned}

Above shows 10 steps for the supposed claim of ln(2)=0\ln(2) = 0 . How many of these steps are incorrect?

Adapted from Simon Singh's blog.
Scene is at 6 minute 46 seconds of the episode Sky Police from The Simpsons.
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