Doing some tests it converges for 1, 2, 4, 5 after about 10 terms, it fails to converge for 6. Still, I'm enthralled this works.
Doing some more tests, if I do the same process over again, but take the little taylor series of ln(x) centered about 0 instead of 1 (we can do this because negative infinity is the identity of \( \bigtriangleup \)) I find the normal taylor series of ln(x): \( ln(x) = \sum_{n=1}^{\infty} (x-1)^n \frac{(-1)^{n+1}}{n} \)
so this leads me to believe that if we take the little taylor series centered about K and then do the same process to get a normal power series expression, and we let K tend to infinity, we will have an ever growing radius of convergence. That is since doing this method centered about 1 has a bigger radius of convergence than doing this method centered about 0.
Doing some more tests, if I do the same process over again, but take the little taylor series of ln(x) centered about 0 instead of 1 (we can do this because negative infinity is the identity of \( \bigtriangleup \)) I find the normal taylor series of ln(x): \( ln(x) = \sum_{n=1}^{\infty} (x-1)^n \frac{(-1)^{n+1}}{n} \)
so this leads me to believe that if we take the little taylor series centered about K and then do the same process to get a normal power series expression, and we let K tend to infinity, we will have an ever growing radius of convergence. That is since doing this method centered about 1 has a bigger radius of convergence than doing this method centered about 0.
