Can e Be Defined as a Limit with n Approaching Negative Infinity?

[PFuser1232]

I am familiar with the fact that the number e can be defined several ways. One particularly interesting definition is the one based on limits, namely:
e = lim_{n $\rightarrow$ ∞} (1 + $\frac{1}{n}$)ⁿ
My question is: wouldn't it be equally true to express e as the limit of the expression above as n goes to NEGATIVE infinity?

 

[Nick O]

According to WolframAlpha, that is quite correct. The proof below is for positive infinity (WA won't show the step-by-step solution for the negative version unless I buy a pro subscription), but the steps are equally valid for negative infinity in this case.

(Open the image in a new tab if you see it with a black background.)

 

[statdad]

"My question is: wouldn't it be equally true to express e as the limit of the expression above as n goes to NEGATIVE infinity?"

Yes it is. The verification is an exercise in algebra and exponent chasing.

If $m$ is a negative integer then $m = -n$ for a positive integer $n$. For $n > 1$ this gives

$$\begin{align*} \left(1 + \frac 1 m\right)^m & = \left(1 - \frac 1 n \right)^{-n} = \left(\frac{n - 1}n\right)^{-n} \\<br /> & = \left(\frac n {n-1}\right)^n = \left(\frac n {n-1}\right) \cdot \left(\frac n {n-1}\right)^{n-1} \\<br /> & = \left(\frac n {n-1} \right) \cdot \left(\frac{n-1}{n-1} + \frac 1 {n-1}\right)^{n-1} \\<br /> & = \left(\frac n {n-1} \right) \cdot \left( 1 + \frac 1 {n-1}\right)^{n-1} = A_n \cdot B_n \text{ (say)}<br /> \end{align*}$$

Note that $\lim_{m \to -\infty} \left(1 + \frac 1 m\right)^m$ equals $\lim_{n \to \infty} \left(1 - \frac 1 n \right)^{-n}$

Since
$$\begin{align*}<br /> \lim_{n \to \infty} \left( \frac n {n-1}\right) & = 1 \\<br /> \text{and}\\<br /> \lim_{n \to \infty} \left(1 + \frac 1 {n-1}\right)^{n-1} & = \lim_{n \to \infty} \left(1 + \frac 1 n%<br /> \right)^n = e<br /> \end{align*}$$

putting everything together gives

$$\lim_{m \to -\infty} \left(1 + \frac 1 m\right)^m = \lim_{n \to \infty} \left(1 - \frac 1 n\right)^{-n} = e$$

 

[PFuser1232]

"My question is: wouldn't it be equally true to express e as the limit of the expression above as n goes to NEGATIVE infinity?"

Yes it is. The verification is an exercise in algebra and exponent chasing.

If $m$ is a negative integer then $m = -n$ for a positive integer $n$. For $n > 1$ this gives

$$\begin{align*} \left(1 + \frac 1 m\right)^m & = \left(1 - \frac 1 n \right)^{-n} = \left(\frac{n - 1}n\right)^{-n} \\<br /> & = \left(\frac n {n-1}\right)^n = \left(\frac n {n-1}\right) \cdot \left(\frac n {n-1}\right)^{n-1} \\<br /> & = \left(\frac n {n-1} \right) \cdot \left(\frac{n-1}{n-1} + \frac 1 {n-1}\right)^{n-1} \\<br /> & = \left(\frac n {n-1} \right) \cdot \left( 1 + \frac 1 {n-1}\right)^{n-1} = A_n \cdot B_n \text{ (say)}<br /> \end{align*}$$

Note that $\lim_{m \to -\infty} \left(1 + \frac 1 m\right)^m$ equals $\lim_{n \to \infty} \left(1 - \frac 1 n \right)^{-n}$

Since
$$\begin{align*}<br /> \lim_{n \to \infty} \left( \frac n {n-1}\right) & = 1 \\<br /> \text{and}\\<br /> \lim_{n \to \infty} \left(1 + \frac 1 {n-1}\right)^{n-1} & = \lim_{n \to \infty} \left(1 + \frac 1 n%<br /> \right)^n = e<br /> \end{align*}$$

putting everything together gives

$$\lim_{m \to -\infty} \left(1 + \frac 1 m\right)^m = \lim_{n \to \infty} \left(1 - \frac 1 n\right)^{-n} = e$$

But why should we consider the constraint n>1? Isn't it enough to say n>0?

 

[statdad]

"But why should we consider the constraint n>1? Isn't it enough to say n>0?"

For the limit it doesn't matter. But for the steps in my approach to work I need $n > 1$ because
of the introduction of the denominators of $n - 1$ .