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!
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.