asymptotic of a geometric like series

Question

I am looking for an asymptotic of $\sum_{n\ge r}2^{-3n+O(\sqrt n)}$ when $r\to +\infty$. Any idea how to obtain such an asymtotic? Thanks in advance

Answer 1

Here is a rough attempt at turning my comment into a proof:

The term $O(\sqrt{n})$ is an abuse of notation; what should be written at that stage is a specific function whose growth is dominated by $\sqrt{n}$ in magnitude in the very specific way given in the Landau notation definitions. Convince yourself that this means that whatever function $f(n)$ should be in place of that $O(\sqrt{n})$ actually obeys the following compound inequality, for some positive real constant $k$ and for all sufficiently large $n$:

$$ -k\sqrt{n}\leq f(n)\leq k\sqrt{n} $$

Then what can we say about the exponent $-3n+O(\sqrt{n})$? This is actually bounded by the compound inequality:

$$ -3n\left(1+\frac{k}{\sqrt{n}}\right)\leq-3n-k\sqrt{n}\leq -3n+O(\sqrt{n})\leq -3n+ k\sqrt{n}\leq-3n\left(1-\frac{k}{\sqrt{n}}\right) $$

Thus, depending on the constant $k$ (which is fixed with respect to the limit we are about to take) and the variable $r$, if we know that $n\geq r$ we can say that, for $\epsilon=\frac{k}{\sqrt{r}}$, the follwoing compound inequality holds:

$$ -3n\left(1+\epsilon\right)\leq -3n+O(\sqrt{n})\leq-3n\left(1-\epsilon\right) $$

But this implies that $\sum\limits_{n\geq r}2^{-3n\left(1+\epsilon\right)} \leq \sum\limits_{n\geq r}2^{-3n+O(\sqrt{n})} \leq \sum\limits_{n\geq r}2^{-3n\left(1-\epsilon\right)}$, for the sufficiently large $n$ discussed earlier. These upper and lower bounds are geometric series, and thus a closed form can be written down for them, which will be purely in terms of $k$, a constant that was originally obfuscated in the $O(\sqrt{n})$, and the variable $r$, which will decide the asymptotic behavior of the sum.