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How can I upper-bound the following function:

$$f(n;a)=\sum_{k=0}^{n-1}\frac{(n-a\sqrt{n})^k}{k!}$$

where $0<a<\sqrt{n}$ is a constant. Since it's a partial sum of exponential series, a trivial upper bound is $e^{n-a\sqrt{n}}$. However, as in this question, the partial sum does not reach that bound because the series gets cut off just as the terms get large.

The answer to the aforementioned question uses the Taylor series with the remainder to derive an approximation. However, I am having trouble writing down the remainder function $R_n(n-a\sqrt{n})=\int_0^{n-a\sqrt{n}}\frac{(n-a\sqrt{n}-t)^n}{n!}f^{n+1}(t)dt$, since $f^{n+1}(t)=\left.\frac{\partial^{n+1}}{\partial x^n}e^{x-a\sqrt{x}}\right|_{t=x}$ is a mess.

Does anyone have any ideas?

ADDENDUM

This post has the following bound:

$$f(n;a)\leq e^{n-a\sqrt{n}}-\frac{(n-a\sqrt{n})^n}{n!}$$

However, as the original poster says in his last comment, this bound is too lose, and I agree -- the improvement from the trivial bound seems to be the subtraction of at most a constant, per application of Stirling's approximation). Is there anything better?

Community
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M.B.M.
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1 Answers1

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I'm going to change your expression to $$ \sum_{k=0}^{\mu+a\sqrt \mu} \frac{\mu^k}{k!}. $$ This is morally the same as your function (take $\mu = n-a\sqrt n$, and it's nearly exactly the same). The idea is that the bulk of the mass of the exponential series for $e^\mu$ happens within order of magnitude $\sqrt \mu$ of $\mu$, and the distribution of that mass is roughly normal. In other words, we should have roughly $$ \sum_{k=0}^{\mu+a\sqrt \mu} \frac{\mu^k}{k!} \approx e^\mu \cdot \frac1{\sqrt{2\pi}} \int_{-\infty}^a e^{-t^2/2}\,dt. $$

To make this quantitatively precise, one can use the Berry-Esseen theorem; see this page (and the one of which it is an appendix) for one exposition.

Greg Martin
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