What uniform bounds are there for $|(1-y/n)^n-\exp(-y)|$ over $n=n_0,n_0 +1,...$ and $\ y\in [0,C]$?

  • 1
    $\begingroup$ I guess it can be estimating by $e^{-y}y^{2}/n$ $\endgroup$
    – openspace
    Dec 14, 2017 at 15:56

1 Answer 1


It should be straightforward to get some sort of bound, maybe not a very good one:

First, it's clear from MVT that $$\left|e^t-e^s\right|\le |s-t|\quad(s,t\le 0).$$

And if $0\le\alpha\le1/2$ then $$-\alpha-\log(1-\alpha)=\int_{1-\alpha}^1(-1+1/t)\,dt,$$so $$0\le-\alpha-\log(1-\alpha)\le\alpha\left(\frac1{1-\alpha}-1\right)=\frac{\alpha^2}{1-\alpha}\le\frac{\alpha^2}2.$$

Now suppose $0\le y\le C$ and $n\ge 2C$, so $0\le y/n\le 1/2$. Then $$\left|-y-n\log(1-y/n)\right|=n\left|-y/n-\log(1-y/n)\right|\le n\frac{(y/n)^2}{2}\le\frac{y^2}{4C}\le\frac{C}4.$$So the inequality for the exponential at the start shows that $$\left|e^{-y}-(1-y/n)^n\right|\le\frac{C}4.$$

That's a bound that's uniform in $y$ and $n$ (subject to the constraints above). And come to think of it it's utterly moronic; the trivial bound$$\left|e^{-y}-(1-y/n)^n\right|\le2$$ is better for $C>8$. If you want a bound that's uniform in $y$ and tends to $0$ as $n\to\infty$ note that what's above shows that $|-y-n\log(1-y/n)|\le y^2/(2n),$ hence $$\left|e^{-y}-(1-y/n)^n\right|\le\frac{y^2}{2n}\le\frac{C^2}{2n}$$for $0\le y\le C$ and $n>2C$.


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