Upper bounding simple infinite sum

Question

I seek to find a $c, \beta > 0$ independent of (uniformly bounding) $s, t \in [0,1]$ such that $$\sum_{n=2}^\infty (-1)^n \frac{2^n+1}{n!}|t-s|^n \le c|t-s|^{1+\beta}$$ for all $s, t \in [0,1]$. I know LHS converges for each $s, t$ because $2^n \ll n!$ asymptotically, but I'm not sure how to get a $c, \beta$ that prove the bound for all $s, t$.

Answer 1

First, see that we can take $|t-s|\to 0$ by simply choosing $c$ to be large enough. That is, $|t-s|^{0.99} < c|t-s|^{0.001}$ for large enough $c$. So we'll work with $|t-s| \to 0$. Now call $x = |t-s|$ for clarity, and see that $$\sum_{n=2}^\infty (-1)^n\frac{2^n+1}{n!}x^n \le \sum_{n=2}^\infty \frac{2^n+1}{n!}x^n$$ Where now dividing both sides by $x^{1+\beta}$ we seek to prove we can find a $\beta$ such that $$\sum_{n=2}^\infty \frac{2^n+1}{n!}x^{n-1-\beta} \le c$$ for large enough $c$. Of course, any $|\beta| <1$ gives each term is $x^n$ for $n>0$. Then taking the limit $x \to 0$ at the beginning means that this inequality is clearly true as the left hand sides vanishes asymptotically. Thus we know $|\beta|<1$ suffices for large enough $c$, and we can conclude, as required.

Answer 2

Since $|t-s|\leq 1$, it follows that for any $n\geq 2$, $|t-s|^n\leq |t-s|^2$, hence your sum is bounded above by $$ \sum_{n= 2}^\infty \frac{2^n+1}{n!} |t-s|^2 = (e^2+e-5) \, |t-s|^2 $$