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Use monotone convergence theorem to prove that $e(x):=\lim_{n\to\infty}(1+x/n)^n$ exists for all real $x$. Then show that $e(-x):=\lim_{n\to\infty}(1-x/n)^n=1/e(x)$.

I'm struggling to use monotone convergence theorem with this; obviously $e(x)$ is monotone when $x>0$, but it's not when $-n<x<0$. So now what?

Further, what makes the second part so difficult is that we can't use any derivatives, logarithms, prior knowledge of $e^x$, etc. We have to do this using just the Bernoulli inequality and finite geometric series, or the regular definition of convergence. I'm just not making any progress and I don't understand how I'm supposed to do this.

How can you show both of these things?

amWhy
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EllipticalInitial
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2 Answers2

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In THIS ANSWER, I showed using Bernoulli's Inequality that the sequence $e_n(x)$, defined by $e_n(x)=\left(1+\frac xn\right)^n$ increases monotonically for $x>-n$.

Moreover, we have from the Binomial Theorem that

$$\begin{align} |e_n(x)|&=\left|\sum_{k=0}^n\binom{n}{k}\left(\frac xn\right)^k\right|\\\\ &\le\sum_{k=0}^n \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right) \frac{|x|^k}{k!}\\\\ &\le \sum_{k=0}^n \frac{|x|^k}{k!} \end{align}$$

It is easy to show (e.g., apply the ratio test) that the series $\sum_{k=0}^\infty \frac{|x|^k}{k!}$ converges for all $|x|$. Hence, we have

$$0\le \sum_{k=0}^n \frac{|x|^k}{k!}\le \sum_{k=0}^\infty \frac{|x|^k}{k!}<\infty$$

Inasmuch as $e_n(x)$ is monotonically increasing and bounded above, then it converges, as was to be shown!

Let $e(x)=\lim_{n\to \infty}e_n(x)$. Then, noting that

$$1\ge e_n(x)e_n(-x)=\left(1-\frac{x^2}{n^2}\right)^n\overbrace{\ge}^{\text{Bernoulli's Inequality}} 1-\frac{x^2}{n}$$

and applying the squeeze theorem, we find that

$$\lim_{n\to \infty}e_n(x)e_n(-x)=1$$

Therefore, we conclude that

$$e(x)e(-x)=1$$

and we are done!

Botond
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Mark Viola
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In the following, we will assume that $n\in\mathbb{Z}^+$ and $x\gt-n$.

Generalizing the result from this answer, using Bernoulli's Inequality, which is proven in that answer, $$ \begin{align} \frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^n} &=\frac{n+x}{n}\,\left(\frac{n+x+1}{n+1}\frac{n}{n+x}\right)^{n+1}\\ &=\frac{n+x}{n}\,\left(1-\frac{x}{(n+x)(x+1)}\right)^{n+1}\\[6pt] &\ge\frac{n+x}{n}\,\left(1-\frac{x}{n+x}\right)\\[12pt] &=1\tag1 \end{align} $$ That is, $\left(1+\frac xn\right)^n$ is an increasing sequence in $n$.

Using the extension of Bernoulli's Inequality to rational exponents proven in this answer, and then to real exponents by continuity, we get the generalization $$ \begin{align} \frac{\left(1+\frac{x}{n}\right)^{n+x}}{\left(1+\frac{x}{n+1}\right)^{n+x+1}} &=\frac{n}{n+x}\left(\frac{n+x}{n}\frac{n+1}{n+x+1}\right)^{n+x+1}\\ &=\frac{n}{n+x}\left(1+\frac{x}{n(n+x+1)}\right)^{n+x+1}\\[6pt] &\ge\frac{n}{n+x}\left(1+\frac{x}{n}\right)\\[12pt] &=1\tag2 \end{align} $$ That is, $\left(1+\frac xn\right)^{n+x}$ is a decreasing sequence in $n$.

Since $$ \left(1+\frac xn\right)^x\ \stackrel{n\to\infty}\searrow\ 1\tag3 $$ we have $$ \left(1+\frac xn\right)^n\le\left(1+\frac xn\right)^{n+x}\tag4 $$ where $(1)$ shows the left-hand side increases in $n$, $(2)$ shows the right-hand side decreases in $n$, and $(3)$ shows that the two sides converge.

Thus, we have the existence of $$ \exp(x)=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag5 $$

If we choose $n$ large enough so that $\left|\frac{xy}{n+x+y}\right|\lt1$, then Bernoulli's Inequality and the fact that for $|x|\lt1$, $1+x\le\frac1{1-x}$ give $$ 1+\frac{xy}{n+x+y} \le\left(1+\frac{xy}{n(n+x+y)}\right)^n \le\frac1{1-\frac{xy}{n+x+y}}\tag6 $$ Thus, the Squeeze Theorem yields $$ \lim_{n\to\infty}\left(1+\frac{xy}{n(n+x+y)}\right)^n=1\tag7 $$ Therefore, we get the formula $$ \begin{align} \exp(x)\exp(y) &=\lim_{n\to\infty}\left(1+\frac xn\right)^n\lim_{n\to\infty}\left(1+\frac yn\right)^n\\ &=\lim_{n\to\infty}\left(1+\frac{x+y}n+\frac{xy}{n^2}\right)^n\\ &=\lim_{n\to\infty}\left(1+\frac{x+y}n\right)^n\lim_{n\to\infty}\left(1+\frac{xy}{n(n+x+y)}\right)^n\\[3pt] &=\exp(x+y)\tag8 \end{align} $$

robjohn
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