$f(x_0) = \sum\limits_{n=1}^\infty (-1)^{n+1} \cdot \frac{x^n}{n} + r_n$
That should say
$$f(x)=\sum_{n=1}^k (-1)^{n+1} \cdot \frac{x^n}{n} + r_k(x),$$
where $r_k$ is the error term of the $k^\text{th}$ partial sum. You want to use estimates to show that the error term goes to $0$ as $k$ goes to $\infty$, which will justify convergence of the series to $f(x)=\log(1+x)$.
Edit: I've struck through part of my answer that relied on a wrong estimate of the derivatives, as pointed out by Robert Pollack. With the missing $k!$ term, the estimate only works on $[-\frac{1}{2},1)$.
Added: To make this answer a little more useful, I decided to look up a correct method. Spivak in his book Calculus (3rd Edition, page 423) uses the formula
$$\frac{1}{1+t}=1-t+t^2-\cdots+(-1)^{n-1}t^{n-1}+\frac{(-1)^nt^n}{1+t}$$
in order to write the remainder as $r_n(x)=(-1)^n\int_0^x\frac{t^n}{1+t}dt$. The estimate $\int_0^x\frac{t^n}{t+1}dt\leq\int_0^xt^ndt=\frac{x^{n+1}}{n+1}$ holds when $x\geq0$, and the harder estimate
$\left|\int_0^x\frac{t^n}{1+t}\right|\leq\frac{|x|^{n+1}}{(1+x)(n+1)}$, when $-1\lt x\leq0$,
is given as Problem 11 on page 430. Combining these, you can show that the sequence of remainders converges uniformly to $0$ on $[-r,1]$ for each $r\in(0,1)$.
Lagrange's form of the error term can be used to do this. The estimates, which follow from Taylor's theorem, are also found on Wikipedia. In this case, if $0\lt r\lt 1$, then $|f^{k+1}(x)|\leq \frac{1}{(1-r)^{k+1}}$ whenever $x\geq-r$, so you have the estimate $|r_k(x)|\leq \frac{r^{k+1}}{(1-r)^{k+1}}\frac{1}{(k+1)!}$ for all $x$ in $(-r,r)$, which you can show goes to $0$ (because (k+1)! grows faster than the exponential function $\left(\frac{r}{(1-r)}\right)^{k+1}$), thus showing that the series converges uniformly to $\log(1+x)$ on $(-r,r)$. Since $r$ was arbitrary, this shows that the series converges on $(-1,1)$, and the convergence is uniform on compact subintervals.
The textbook "An introduction to numerical methods and analysis" by James F. Epperson, 2nd edition, page 3:
$$
\cos x = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + \dotsm + \frac{(-1)^n}{(2n)!}x^{2n} + \frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2} \cos \xi_x
$$
So the answer is yes, because the first term for remainder is zero then you can go for the next n.
Best Answer
I see that this is a pretty old question, but here goes anyway.
The main problem with a geometric approach is that we are often dealing with very high order derivatives. Just by looking at a graph we can easily get a sense of such geometric interpretations as value, gradient and concavity - corresponding respectively to $f^{(0)}(x)$, $f^{(1)}(x)$ and $f^{(2)}(x)$ - but after that it starts to become a struggle to interpret function behaviour visually.
Having said that, if we choose $f(x)=e^x$, for which $f^{(k)}(x) = e^x$ for all $k$, we can produce something of a geometric interpretation of the Lagrange error term, especially if we start off with low degree Taylor polynomials for the approximation and incrementally incorporate more terms from the series into the polynomial approximation.
We know that:
$$\begin{align} f(b) = \sum_{k=0}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}&= \sum_{k=0}^n\frac{f^{(k)}(a)(b - a)^k}{k!}+\sum_{k=n+1}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}\\\\ &= T_n(b:a) + R_n(b:a) \end{align}$$
And that:
$$\left(\exists c \in ]a, b[\right)\left(\frac{f^{(n+1)}(c)(b-a)^{n+1}}{(n+1)!}=R_n(b:a)=\sum_{k=n+1}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}\right)$$
Now, for example, let's say that we want to use a Taylor series of $f(x) = e^x$ about $a = 0$ to estimate $f(b = 2)$.
If we make the $n=0$ assertion (in other words, we say that $f(b)\approx f^{(0)}(a)$ independent of $b$, which generally is only a good idea for $b$ very close to $a$), we are effectively saying that there is some $c$ in the region $]a, b[$ such that:
$$\frac{f^{(0+1)}(c)(b-a)^{0+1}}{(0+1)!}=e^b - e^a$$
In this particular example, where we have specified the values, we can calculate $c$:
$$2e^c=e^2 - 1\\c = \ln\left(\frac{e^2 - 1}{2}\right)\approx 1.16$$
With $c$ and $f(x)$ translated down by twice the gradient of $f$ at $c$ (which, since we're using $f(x)=e^x$, is twice $f(c)$) shown:
Similarly, if we make the $n=1$ assertion (in other words we say that $f(b)\approx f^{(0)}(a)+ f^{(1)}(a)(b - a)$) we are effectively saying that there is some $c$ in the region $]a, b[$ such that:
$$\frac{f^{(1+1)}(c)(b-a)^{(1 + 1)}}{(1+1)!}=e^b - (e^a + e^a(b - a))$$
Since we have specified all values we can again calculate $c$, which, together with the graph of $f(x)$ translated down by twice the concavity of $f$ at $c$ (which, since we're using $f(x) = e^x$, is twice the value of $f$ at $c$) is illustrated:
Of course, we can continue on similarly from here (though we have to now pay a little more attention to the factorials in the denominators), incorporating successively more terms from the Taylor series into the Taylor polynomial. The nice thing about using $f(x)=e^x$ is that we will always be able to interpret derivatives evaluated at $c$ as the height of the curve above the $x$ axis at $c$, which would not be the case with other functions.
While a geometric interpretation of the Lagrange error term would be a lot more complicated with other functions, I found that when I could make sense of what was happening in the simple $e^x$ case, the Lagrange error term made a lot more sense in general.