Taylor series of hyperbolic tangent function tanh(x) For the analytic hyperbolic tangent function $\tanh(x)$, denote the Taylor (Maclaurin) series around $x = x_0$ as $g(x;x_0)$ as follow
\begin{equation}
 g(x; x_0) = \sum\limits_{n=0}^\infty\left.\frac{\tanh^{(n)}(x)}{n!}\right|_{x=x_0}(x-x_0)^n
\end{equation}
Question. When $0 < x_0-\epsilon < x < x_0$, $\epsilon > 0$ (for example $\epsilon = 1$), whether $\tanh(x) > g(x;x_0)$ or $\tanh(x) < g(x;x_0)$ when $x \in (x_0-\epsilon,x_0)$? 
For example. Let $x_0 = 2$ and $\epsilon=1$, assume that we knew all-order derivatives of $tanh(x)$ at $x = x_0 = 2$, but we do not know derivatives of $tanh(x)$ at other points. Use $g(x;2)$ to estimate the value of $tanh(x)$ when $x=1.5 \in (1,2)$. We want to know whether this estimated value $g(1.5;2)$ is larger or less than the actual value of $tanh(1.5)$.
 A: What you have written is not the Taylor series for $\tanh$ around $x_0$, the Taylor series requires you to take the derivates at the point you expand around. If it were the Taylor series you would have had
$$\tanh(x) = \sum_0^\infty{\tanh^{(n)}(x_0)\over n!}(x-x_0)^n$$
as long as at least $|x-x_0| < \sqrt{x_0^2+\pi^2/4}$ (if on the other hand $|x-x_0|>\sqrt{x_0^2+\pi^2/4}$ the series diverges). Certainly if $0<x<x_0$ we have that $|x-x_0| < |x_0| < \sqrt{x_0^2+\pi^2/4}$ so then the series converges (to $\tanh(x)$).
As you wrote $g(x;x_0)$ it resembles the Taylor expansion around $x$ instead except you've changed the sign of the $x_0-x$. It's indeed a Taylor expansion around $x$, but it's not necessarily $\tanh$ that is expanded. To see what you've expanded you can use even/odd decomposition of the function. Let
$$f(x_0) = \sum_0^\infty{\tan^{(n)}(x)\over n!}(x-x_0)^n$$
And consider $f_e(x_0) = (f(x_0)+f(-x_0))/2$ and $f_o(x_0) = (f(x_0)-f(-x_0))/2$, we see that $f_e$ is even and $f_o$ is odd and $f=f_e+f_o$, that is $f_e$ is the even part and $f_o$ is the odd part of $f$. Now we have:
$$f_e(x_0) = \sum_0^\infty{\tan^{(2n)}(x)\over (2n)!}(x-x_0)^{2n} = \sum_0^\infty{\tan^{(2n)}(x)\over (2n)!}(x_0-x)^{2n} = \tanh_e(x_0)$$
$$f_o(x_0) = \sum_0^\infty{\tan^{(2n+1)}(x)\over (2n+1)!}(x-x_0)^{2n+1} = -\sum_0^\infty{\tan^{(2n+1)}(x)\over (2n+1)!}(x_0-x)^{2n+1}=-\tanh_o(x_0)$$
Where $\tanh_e$ is the even part and $\tanh_o$ is the odd part of $\tanh$, so we have:
$$f(x_0) = \tanh_e(x_0) - \tanh_o(x_0) = {\tanh(x_0) + \tanh(-x_0)\over2}-{\tanh(x_0)-\tanh(-x_0)\over 2} = \tanh(-x_0)$$
This means that $g(x;x_0) = \tanh(-x_0)$ at least if $|x_0-x|<\pi/2$. This means that especially that since $x_0>0$ that $g(x;x_0)$ is negative and since $x>0$ that $\tanh(x)>0$ so you have $\tanh(x) > g(x;x_0)$.
Note that you could have found what $g(x; x_0)$ is by noting that $\tanh$ is odd so $\tanh^{(n)}(x) = -\tanh^{(n)}(-x)(-1)^n$ so $\tanh^{(n)}(x_0)(x-x_0) = -\tanh^{(n)}(-x)(x_0-x)^n$ so the series is that of $-\tanh(x_0)$ around $-x$.
