I want to take the series for arcosh(x), square the first part of all of the terms, subtract 1 from the denominator of the second part of all the terms, subtract the "ln2x", add 1 to the entire series, and multiply the entire series by 2pi. How do these adjustments alter the original arcosh(x) function? Actually, if anyone can tell me how just one of these adjustments, say the subtraction of ln2, affects the original function, I will be much indebted.

$\displaystyle \ arcosh(x)= \ln 2x - \sum_{n=1}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-2n}} {(2n)} , \qquad x > 1 $

i.e.

$\displaystyle arcosh(x)= \ln 2x - \left( \left( \frac {1} {2} \right) \frac {x^{-2}} {2} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{-4}} {4} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{-6}} {6} +\cdots \right)$

The adjustments I want to make:

$\displaystyle 2\pi \left( \ln 2x - \ln 2x + 1 - \sum_{n=1}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right)^2 \frac {x^{-2n}} {(2n-1)} \right) , \qquad x > 1 $