Series for ln(x)?

[teal hull]

So everyone knows about how ln(1 + x) = sum(n = 0, inf) (-1)^n/(n + 1) x^(n + 1), for -1 < x <= 1. I was playing around with this series, and I think I've discovered a general series form for the natural log of any real number greater than or equal to 1/2; ln(x) = sum(n = 1, inf) (x - 1)^n/(nx^n). I'd appreciate if someone could tell me if I made a mistake somewhere.

The general argument runs thus: ln(x) = -ln(1/x) = -ln(1 + (1/x - 1)) = -ln(1 - (1 - 1/x)) = -ln(1 - (x - 1)/x) = - sum(n = 0, inf) (-1)^n/(n + 1) * (-(x - 1)/x)^(n + 1) = sum(n = 0, inf) (-1)^(n + 1)/(n + 1) * (-1)^(n + 1) * ((x - 1)/x)^(n + 1) = sum(n = 0, inf) ((-1)^2)^(n + 1)/(n + 1) * ((x - 1)/x)^(n + 1) = sum(n = 0, inf) (x - 1)^(n + 1)/((n + 1)x^(n + 1) = sum(n = 1, inf) 1/n * ((x - 1)/x)^n
For the range of convergence, we start with the range of convergence of ln(1 + t): -1 < t <= 1 t = -(x - 1)/x -1 < -(x - 1)/x <= 1 1 > (x - 1)/x >= -1
Since x > 0 by the domain of ln(x): x > x - 1 >= -x
Now, x > x - 1 is trivially true for real numbers, so: x - 1 >= -x 2x - 1 >= 0 2x >= 1 x >= 1/2

So that's my work. Does this actually result in a series for ln(x) for arbitrarily large x?

[golden torrent]

Yeah, it does. Here's a more general series that I've discovered: it converges for any x > 0.
Basically, it exploits the property ln(1/x) = -ln(x).

The convergence is still pretty slow, though, as usual.

This series, derived from the series of artanh(x), converges faster (and also for all z > 0).

[teal hull]

Do you mean z =/= 0? Since z is usually used for complex numbers and complex numbers aren't well-ordered.

[golden torrent]

Oh, nah, that's just a series I quickly googled when I remembered about it. z is a positive real number here.

[teal hull]

Weird that they'd use z and not x then.