Formula to $\ln$ that holds on interval $x \geq 1$ In the Wikipedia we can find two formulas using power series to $\ln(x)$,
but I would like a formula that holds on the interval $x \geq 1$ (or is possible to calculate $\ln(x)$ to $x \geq 1$ with the same formula).
Is there one?
In other words, a formula like,
$$\ln(x)=\sum_{i=0}^{\infty}a_i x^i$$
The wikipedia formulas are:
$$\ln(x+1)=\sum_{i=1}^{\infty}\frac{(-1)^{i-1}x^i}{i}\text{ for }1 < x \leq1$$
and
$$\ln\left(\frac{x}{x-1}\right)=\sum_{i=1}^{\infty}\frac{1}{i x^i}$$
PS.: Is important to me to use power series, but, any help are welcome.
 A: To add to what was written by Gerry Myerson, if $|w|<1$, we have by the usual power series expansion for $\log(1+w)$,
$$\log(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}-\frac{w^4}{4}+\cdots.\tag{$1$}$$
Replacing $w$ by $-w$, we have
$$\log(1-w)=-w-\frac{w^2}{2}-\frac{w^3}{3}-\frac{w^4}{4}-\cdots.\tag{$2$}$$
Subtracting $(2)$ from $(1)$, we obtain
$$\log(1+w)-\log(1-w)=\log\left(\frac{1+w}{1-w}\right)=2w+2\frac{w^3}{3}+2\frac{w^5}{5}+\cdots.\tag{$3$}$$
For any $x>1$, there is a $w$ between $0$ and $1$ such that $x=\frac{1+w}{1-w}$. We can solve for $w$ explicitly in terms of $x$, obtaining $w=\frac{x-1}{x+1}$.
Thus we get an expansion for $\log x$ in terms of powers of $\frac{x-1}{x+1}$. This is not a power series expansion of $\ln x$, but it can be useful.  For large $x$, the number $w$ is close to $1$, so the convergence is relatively slow. But for smallish $x$, there is usefully fast convergence. Euler, among others, used the series $(3)$ for computations.
Take for example $x=2$. The ordinary series for $\log(1+x)$ converges when $x=1$, but too slowly for practical use. However, in this case $w$ turns out to be $1/3$, and the series $(3)$ converges quite fast. For $x=3$, we get $w=1/2$, and again we get usefully fast convergence.
A: At http://www.efunda.com/math/exp_log/series_exp.cfm you will find $$\log x=2\sum_1^{\infty}{1\over2n-1}\left({x-1\over x+1}\right)^{2n-1}$$ claimed to be valid for $x\gt0$, and several other formulas you might enjoy. 
A: Another alternative, that works for $x > -\dfrac 1 2 $ is
$$\log (1+x) =\sum_{n>0} \frac 1 n\left( \frac{x}{x+1}\right)^n$$
This is trivially obtained by setting $u=\dfrac{x}{x+1}$ in
$$-\log(1-u)=\sum_{n >0}\frac {u^n} n $$
or by showing 
$$\int_0^x \left( \frac{t}{t+1}\right)^{n+1}\frac{dt}t=\log(1+x)-\sum_{k \leq n} \frac 1 k\left( \frac{x}{x+1}\right)^k$$
Then since
$$\left|\log(1+x)-\sum_{k \leq n} \frac 1 k\left( \frac{x}{x+1}\right)^k\right|=\left| \int_0^x \left( \frac{t}{t+1}\right)^{n+1}\frac{dt}t \right|<\epsilon$$
for sufficiently large $n$ and $x > -\dfrac 1 2 $ the equality follows.
