Upper bounding using negative exponentials

We know that for a fixed $x>0$, the convergence of $$\left(1+\frac{x}{n}\right)^n\rightarrow e^x$$ is from below and so we have the following bound for any $n\in\mathbb{N}$:

$$\left(1+\frac{x}{n}\right)^n\leq e^{x}.$$

However, for a fixed $y<0$, $y=-x$ the convergence of $$\left(1+\frac{y}{n}\right)^n=\left(1-\frac{x}{n}\right)^n\rightarrow e^{-x}$$ is from above and so we certainly can't write $$\left(1-\frac{x}{n}\right)^n\leq e^{-x}.$$

EDIT: No it isn't and yes we can. See answer below.

Is there a well-known way to salvage an upper bound here? Perhaps a bound of the form $$\left(1-\frac{x}{n}\right)^n\leq \phi(x)e^{-x},$$ for some relatively small $\phi(x)$.

Context:

At the moment, for a fixed $n$, I am hoping to upper bound the term $$\left(1-\frac{(n-1)(\sqrt{n}-1)^2}{n^n-1}\right)^{2k},$$

where I have control on $k$. I had mistakenly taken $k=\alpha(n^n-1)$ for some $\alpha>0$ and concluded that

$$\left(1-\frac{(n-1)(\sqrt{n}-1)^2}{n^n-1}\right)^{2k}\leq e^{-2\alpha(n-1)(\sqrt{n}-1)^2},$$

but of course this is incorrect and I need something more along the lines of

$$\left(1-\frac{(n-1)(\sqrt{n}-1)^2}{n^n-1}\right)^{2k}\leq \phi((n-1)(\sqrt{n}-1)^2)e^{-2\alpha(n-1)(\sqrt{n}-1)^2},$$

• @Dr.MV I'm sorry I don't see how that helps... – JP McCarthy Aug 15 '16 at 22:45
• $$\left( 1-\frac{x}{n} \right) ^n \color{blue}{\leq} e^{-x}$$ In this case $e^{-x}$ is the upper bound. How would you make it any closer? I don't understand what you are asking – Yuriy S Aug 16 '16 at 0:47
• @Yuriy S I was mistaken. – JP McCarthy Aug 16 '16 at 6:53

There is a misstatement in the OP. The sequence $e_n=\left(1-\frac xn\right)^n$ is actually an increasing sequence. To see this, we write for $x<n$

\begin{align} \frac{e_{n+1}}{e_n}&=\frac{\left(1-\frac x{n+1}\right)^{n+1}}{\left(1-\frac xn\right)^n}\\\\ &=\left(1-\frac xn\right)\left(1+\frac{x}{(n-x)(n+1)}\right)^{n+1} \tag 1\\\\ &\ge \left(1-\frac xn\right)\left(1+\frac{x}{(n-x)}\right) \tag 2\\\\ &=1 \end{align}

where in going from $(1)$ to $(2)$ we exploited Bernoulli's Inequality.

Therefore, $e_n$ increases monotonically for $x<n$ to $e^{-x}$ and

$$e^{-x}\ge \left(1-\frac xn\right)^n$$

• You're welcome. My pleasure. -Mark – Mark Viola Aug 16 '16 at 14:06
• Yes I was able to use this to get a nice bound... the source of my earlier confusion was for $x>0$ that $(1-x/n)^n\geq (1-x/n)^{n+1}$... – JP McCarthy Aug 16 '16 at 14:12
• That is correct. We have $$\left(1-\frac x{n+1}\right)^{n+1} \ge \left(1-\frac xn\right)^n \ge \left(1-\frac xn\right)^{n+1}$$ – Mark Viola Aug 16 '16 at 14:41