error of Taylor series for $\ln x$ 
Find the Taylor Polynomials for $f(x)=\ln x$ about $a=1$ and give error estimates.
Below is what I've done. There may be some mistakes.

Let $f(x) = \ln x$. Then $f(1) = 0$. $f'(x) = \dfrac{1}{x}$ so $f'(1) = 1$. $f''(x) = -\dfrac{1}{x^2},$ so $f''(1) = -1$. $f^{(3)}(x) = \dfrac{2}{x^3}$ so $f^{(3)}(1) = 2$. From this, we see that $f^{(n)}(1)$, where $n>0,$ has value $(-1)^{n-1}(n-1)!$. Hence the Taylor polynomials $P_{n,1}(x)$ for $f(x)$ about $a=1$ are given by $\displaystyle\sum_{i=1}^n (-1)^{i-1}\dfrac{(x-1)^i}{i}$. By Taylor's Theorem, we have that $f(x) - P_{n,1}(x)=\dfrac{f^{(n+1)}(x_0)}{(n+1)!}(x-1)^{n+1},$ where $1\leq x_0 \leq x$. Hence since $|f^{(n+1)}(x_0)|=|(-1)^n\dfrac{n!}{(x_0)^{n+1}}|=\dfrac{n!}{(x_0)^{n+1}}, n\in\mathbb{N}$ is a decreasing function, for $x_0\geq1$, it has a maximum of $n!$ at $x_0=1$. Thus, the absolute value of the error is given by $\dfrac{n!}{(n+1)!}|(x-1)|^{n+1}=\dfrac{|(x-1)|^{n+1}}{n+1}.$

Edit: I guess this could also be shown using the fact that the error for an alternating series is smaller than the next term and has the same sign.

 A: Do yourself a favor by working with $\ g(u):=\ln(1+u)\, $ -- this would go smoother. Then substitute,
$$ f(x)\,:=\,\ln(x)\, =\, g(x-1) $$
If you were interested in a different simple power series which computes
logarithms clearly faster (the error term would converge to $0$ more
rapidly) then let me know, and I will expand my Answer.

 
        Here we go:

We work in the field of complex numbers $\,\Bbb C\,$ (so much
better than $\,\Bbb R\,$ alone).
While the standard polynomial power series $\,\sum a_k\cdot x^k\,$
are wonderful, the rational power series $\,\sum a_k\!\cdot\!(r(x))^k\,$
(function $r(x)$ being a quotient of two polynomials) form a much larger class hence they offer much more potential.
Let's start with standard two series
$$ \ln(1+x)\,=\,\sum_{n=1}^\infty (-1)^{n-1}\cdot\frac{x^n}n $$
and
$$ \ln(1-x)\,=\,-\sum_{n=1}^\infty \frac{x^n}n $$
The convergence radius is $1$ (in both cases; the second series is
obtained from the first one by substitution of $x$ by $-x$).
Subtract the second series from the first one:
$$ \ln\frac{1+x}{1-x}\,\,=\,\,
    2\cdot\sum_{k=1}^\infty \frac{x^{2\cdot k-1}}{2\cdot k-1} $$
The convergence radius is still $1$ and not more (as illustrated by
$\,x=1.)$ All this, so far, can be found in about all texts on
Mathematical Analysis (or Calculus) But the following useful simple
step is hardly in any of them:
substitute $\,t\,:=\,\frac{1+x}{1-x}\,\,$
so that $\,x\,=\,\frac{t-1}{t+1} $
and   (the main result!)
$$ \ln(t)\,\,=\,\,2\cdot\sum_{k=1}^\infty
   \frac 1{2\cdot k-1}\cdot\left(\frac{t-1}{t+1}\right)^{2\cdot k-1}   $$
Now, look closely (how nice!) the series converges for every
$\,t\in\Bbb C\,$ such that $\,\Re(t)>0,\,$ i.e. in the whole
half-plane of the positive real part!
EXAMPLE 1:
$$ \ln(2)\,=\,2\cdot\sum_{k=1}^\infty
      \frac 1{(2\cdot k-1)\cdot 3^{2\cdot k-1}} $$
EXAMPLE 2:
$$ \ln(10)\,=\,2\cdot\sum_{k=1}^\infty
   \frac 1{2\cdot k-1}\cdot\left(\frac9{11}\right)^{2\cdot k-1} $$
EXAMPLE 3:
$$ \ln(11)\,=\,2\cdot\sum_{k=1}^\infty
   \frac 1{2\cdot k-1}\cdot\left(\frac56\right)^{2\cdot k-1} $$
THEOREM
$$ \ln\left(\frac{s+1}s\right)\,\,=\,\,2\cdot\sum_{k=1}^\infty
      \frac 1{(2\cdot k-1)\cdot(2\cdot s+1)^{2\cdot k-1}}
    \qquad\mbox{whenever}\quad   \Re\left(\frac 1s\right)>-1  $$
Now, we can have a lot of fun by applying the above in the
context of equations like
$$ 2=\left(\frac 43\right)^2\cdot\frac 98\qquad\mbox{or}\qquad
        3=\left(\frac 43\right)^3\cdot\left(\frac 98\right)^2 $$
or
$$ 5\,=\,\left(\frac32\right)^4\cdot\frac{80}{81} $$
and a lot more.

REMARK   It follows from the above main result that

$$ \ln\left(\frac 1t\right)\,=
      \,-\ln(t)\qquad\mbox{whenever}\quad\Re(t)>0 $$
