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i spose we would nee to generate the power series for it first

maybe?

lol
-ln(1+2x) = -2/(1+2x)

2 i think

somewhere i had seen something like this
integration dln(1+2x)/dx
and expand it as power series.

soo, a relevant power series would be:\[ln(\frac{1}{1+2x})=\int\sum (-1)^{n+1}\ 2^{2n}x^{n}dx\]

\[ln(\frac{1}{1+2x})=\sum_{0}^{inf}\frac{(-1)^{n+1}}{n+1}x^{n+1}\]

forgot the 2^n

2^(n+1)

prolly shoulda taken the - with the x :)

all non - ns should be vacated
\[|-2x|\ lim\frac{n}{n+1}=2x\]
\[2x<1;\ x<\frac{1}{2}\]