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