Suppose your series is $f(x) = \sum_{n=0}^\infty a_n x^n$ with the signs of $a_n$ alternating. If $|a_n| r^n$ is decreasing to $0$, this is an alternating series for
$0 < x < r$.
The alternating series bound for the remainder after the $x^n$ term is then $|a_{n+1}| x^{n+1}$. The Lagrange form for the remainder is
$\dfrac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}$, where $0 < c < x$, and to get a bound
we want to maximize $|f^{(n+1)}(c)|$ on this interval.
Now
$$\dfrac{f^{(n+1)}(c)}{(n+1)!} = \sum_{j=n+1}^\infty {j \choose n+1} a_j c^{j-n-1}$$
The bound is the same as the alternating series bound if the maximum occurs at $c=0$.
Now the derivative of this is
$$ \dfrac{f^{(n+2)}(c)}{(n+1)!} = \sum_{j=n+2}^\infty (j-n-1) {j \choose n+1} a_j c^{j-n-2} = (n+2)\sum_{j=n+2}^\infty {j \choose n+2} a_j c^{j-n-2} $$
If it weren't for that ${j \choose n+2}$ factor, this would still be an alternating series, and $f^{(n+2)}(c)$ would have the same sign as
$a_{n+2}$, which is opposite to the sign of $a_{n+1}$ and $f^{(n+1)}(c)$, implying that the maximum is at $c=0$. But that factor can mess things up.
Consider e.g. a series that starts $1 - x + x^2 - x^3 + x^4$, with the remaining terms very small (but still alternating for $0 < x < 1$). You want to estimate
the error in the linear approximation $1 - x$. Then
$$\dfrac{f''(c)}{2} \approx 1 - 3 c + 6 c^2$$
If $ 1/2 < x < 1$, the maximum of this is not at $c=0$ but rather at $c=x$.
The Lagrange bound is then approximately $(1 - 3 x + 6 x^2) x^2$, which is different from the alternating series bound of $x^2$.