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Let $S_N$ denote the taylor polynomial (i.e., the partial sum of the taylor series) (centered at $0$) for a function $g$. Let $g$ be differentiable $N+1$ times on its radius of convergence $(-R, R)$. And let $L_N=g(x)-S_N(x)$.

I'm trying to show that if $0

From a hint, I've shown that if two differentiable functions $A, B$ on $[0,x]$ satisfy $A(0)=B(0)$ and $A'(y)\leq B'(y)$ for $y\in [0,x]$, then $A(y)\leq B(y)$ for $y\in [0,x]$. But how does one use this to prove what I want to show?

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By induction on $N$ we can prove from the hint that if all derivatives of $A$ and $B$ up to $N$th are 0, and $A^{(N+1)}\le B^{(N+1)}$ on $[0,t]$, then $A\le B$ on $[0,t]$. Take $A=L_N$ or $A=-L_N$, $B(x)=\frac{Kx^{N+1}}{(N+1)!}$ and apply this.

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    Won't we need that $L_N$ and $\frac{Kx^{N+1}}{(N+1)!}$ have derivatives equal to $0$ for the first $N$ times? – 2017-01-28
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    Yes, but only in point 0, which is true for $L_N$ by definition of Taylor series and for the fraction by explicit calculation. – 2017-01-28