We call a function $L:(0,\infty)\rightarrow(0,\infty)$ slowly varying if for each $c>0$ one has $\displaystyle \lim_{x\to\infty}\frac{L(cx)}{L(x)}=1$. Can somebody give me a hint why the sum of such functions is also slowly varying?

As the title says. I think this should follow straightforwardly but I can’t find a proof. My random variable of interest $X$ takes values in the non-negative integers. The only other assumption on its distribution is that $E(X)<\infty$. I want to prove: $$\lim_{n\to\infty}n\Pr(X\ge n) = 0.$$ The fact that this should follow is referenced e.g. […]

The Wikipedia page for the Binomial Distribution states the following lower bound, which I suppose can also be generalized as a general Chernoff lower bound. $$\Pr(X \le k) \geq \frac{1}{(n+1)^2} \exp\left(-nD\left(\frac{k}{n}\left|\right|p\right)\right) \quad\quad\mbox{if }p<\frac{k}{n}<1$$ Clearly this is tight up to the $(n+1)^{-2}$ factor. However computationally it seems that $(n+1)^{-1}$ would be tight as well. Even (n+1)^{-.7} […]

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