<s>[INST] Proving difference between two functions is small when $x$ is small... I'm really struggling to prove the following claim and I was wondering whether anyone could help me.

Claim:$$ 0<|x| \leq 10^{-4} \implies \bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12}$$

My attempt at a proof: I have already proven the following inequality,
$$\bigg{|} \frac{e^x-1}{x} - \sum_{k=1}^n \frac{x^{k-1}}{k!} \bigg{|}  \leq \frac{|x|^ne^{|x|}}{(n+1)!}$$
So let $n=3$, then
\begin{align*}
\bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} &= \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})}  \cdot \left( \frac{e^x-1}{x} - \sum_{k=1}^3 \frac{x^{k-1}}{k!}  \right) \bigg{|} \\
&= \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \bigg{|} \frac{e^x-1}{x} - \sum_{k=1}^3 \frac{x^{k-1}}{k!}  \bigg{|} \\
&\leq \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{|x|^3e^{|x|}}{(3+1)!} \\
&\leq \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{(10^{-4})^3 e^{10^{-4}}}{4!} \\
&= 10^{-12} \cdot \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{ e^{10^{-4}}}{4!} 
\end{align*} Hence, it suffices to show that for $0 < |x| \leq 10^{-4}$, 
\begin{align*}
f(x) := \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} < \frac{4!}{e^{10^{-4}}}
\end{align*}
Since $f(x)$ is monotonically decreasing on the intervals $[-10^{-4}, 0)$ and $(0, 10^{-4}]$, we know that for $0 <|x| \leq 10^{-4}$, 
\begin{align*}
f(x) \leq f \left( -10^{-4} \right) \approx  1.0001 < \frac{4!}{e^{10^{-4}}}
\end{align*} Therefore, for $0 < |x| \leq 10^{-4}$,
\begin{align*}
\bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12} 
\end{align*} $\square$
My big problem here is that I don't actually know how to prove my claim that $f(x)$ is monotonically decreasing. Does anyone know how to prove this? 
Or even better, does anyone know of a more direct proof that avoids this problem altogether?
Thanks!
 [/INST] Write $|\frac{x}{e^x-1} - \frac{x}{\sum_{k=1}^3 \frac{x^k}{k!}}| 
 = |x||\frac{\sum_{k=1}^3 \frac{x^k}{k!} - (e^x-1)}{(e^x-1)\sum_{k=1}^3 \frac{x^k}{k!}}| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denominator is obviously increasing in $x$.
So I guess this comes under "direct proof". I think your function $f(x)$ is probably decreasing though.
</s>