Why does a boundedness proof of $\exp(x)$ depend on the sign of $x$?

Question

To prove $\lim\limits_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}$ exists, we prove that the sequence $$f_n=\left(1+\frac{x}{n}\right)^n$$ is bounded and monotonically increasing toward that bound.

Proof Attempt:

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We begin by showing $f_n=\left(1+\frac{x}{n}\right)^n$ is monotonically increasing by looking at the ratio of consecutive terms: \begin{align*} \frac{f_{n+1}}{f_n} &=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n}} \tag{Definition of $f_n$} \\ &=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)^{n}\left(1+\frac{x}{n}\right)} \tag{Multiplication by $\frac{\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)}$} \\ &=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n+1}}\left(1+\frac{x}{n}\right) \tag{Simplify $a^n\cdot a = a^{n+1}$} \\ &=\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a^{n+1}}{b^{n+1}}=\left(\frac{a}{b}\right)^{n+1}$} \\ &=\left(\frac{\frac{n+1+x}{n+1}}{\frac{n+x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Common denominators} \\ &=\left(\frac{n+1+x}{n+1}\cdot \frac{n}{n+x}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$} \\ &=\left(\frac{n^2+n+nx}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Distribute $(n+1+x)n$} \\ &=\left(\frac{n^2+n+nx+x-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Add and subtract $x$} \\ &=\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Factor $n^2+n+nx+x$} \\ &=\left(1+\frac{-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$} \\ &\ge\left(1+\frac{-x}{(n+x)}\right)\left(1+\frac{x}{n}\right) \tag{Bernoulli: $(1+x)^n \ge 1+nx$} \\ &=\left(\frac{n}{n+x}\right)\left(\frac{n+x}{n}\right) \tag{Common denominators} \\ &=1 \tag{Simplify $\frac{a}{b} \cdot \frac{b}{a}=1$} \end{align*} Since $\frac{f_{n+1}}{f_n}>1$, then $f_{n+1}>f_n$, which shows the sequence $f_n$ is monotonically increasing for all $n \in \mathbb{N}$.

Next, we show $f_n=\left(1+\frac{x}{n}\right)^n$ is bounded above. Note that \begin{align*} f_n &=\left(1+\frac{x}{n}\right)^n \tag{Definition of $f_n$} \\ &=\sum_{k=0}^n \binom{n}{k} (1)^{n-k} \left(\frac{x}{n}\right)^{k} \tag{Binomial Theorem} \\ &=1+\frac{n}{1!}\left(\frac{x}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac{x}{n}\right)^3+\cdots+\frac{n!}{n!}\left(\frac{x}{n}\right)^n \\ &=1+\frac{\frac{n}{n}}{1!}x+\frac{\frac{n(n-1)}{n^2}}{2!}x^2+\frac{\frac{n(n-1)(n-2)}{n^3}}{3!}x^3+\cdots+\frac{\frac{n!}{n^n}}{n!}x^n \tag{Simplify}\\ &=1+\frac{1}{1!}x+\frac{\left(1-\frac{1}{n}\right)}{2!}x^2+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}{3!}x^3+\cdots+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)}{n!}x^n \\ & \le 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots+\frac{1}{n!}x^n \tag{$1-\frac{k}{n}<1$} \\ & = \sum_{k=0}^n \frac{1}{k!} x^k \tag{Sigma notation}\\ & \to %\underset{n \to \infty}{\to} \sum_{k=0}^\infty \frac{1}{k!}x^k \tag{as $n \to \infty$} \\ & = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{k!}x^k \tag{$\exists K$, $k>K$ implies $k! \ge (2x)^k$}\\ & \le \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{(2x)^k} x^k \tag{$k! \ge (2x)^k$ implies $\frac{1}{k!} \le \frac{1}{(2x)^k}$}\\ & = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{2^k} \tag{$\frac{1}{(2x)^k}x^k=\frac{1}{2^k x^k}x^k = \frac{1}{2^k}$}\\ &= \sum_{k=0}^K \frac{1}{k!} x^k + \frac{1}{2^K} \tag{Geometric series evaluation} \end{align*} which is finite. Thus, the sequence $f_n$ is bounded. Since it is both monotonically increasing and bounded, it is convergent by the Monotone convergence theorem.

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Is my proof correct? I am suspicious of the step which says "$\rightarrow \sum_{k=0}^n \frac{1}{k!}x^k$", and would like to avoid taking another limit in the middle of the boundedness proof.

I also compared my proof to the following references and saw something worrisome:

• Reference 1 <- Assumes $x\ge 0$ (Why?)

• Reference 2 <- Assumes $x \ge -1$ (Why?)

• Reference 3 <- Considers $x=0$, $x>0$, and $x<0$ separately (Why?)

All of the above proofs either assumed $x>0$ or considered cases where $x>0$ and $x<0$ separately, but I do not know why. In fact, the third reference considers $\left(1-\frac{x}{n}\right)^{-n}$ for $x>0$ (I think this is a typo and should read $x<0$), but I am not sure why the negative exponent is needed (we are talking about a negative value of $x$, not negative $n$.)

I could only find one proof that did not consider different cases on the sign of $x$:

• Bonus reference 4 <- Uses absolute values, but I am not sure why these are necessary either.

I would like to verify my proof and ask 3 questions:

1. Why is it necessary to consider cases $x>0$ and $x<0$ separately? Did any step in my proof implicitly assume that $x>0$? If so, which one?

2. Is there any way to avoid taking a limit in the middle of the boundedness proof?

3. Substituting $n=1$ in my boundedness proof shows $1+x \le \sum_{k=0}^n \frac{1}{k!}x^k$. Does this imply $1+x \le \lim\limits_{n\to \infty} \left(1+\frac{x}{n}\right)^n$, since $f_n$ is an increasing function of $n$? Can this be seen explicitly, or would that require a separate proof?

Thank you.

Answer 1

Let's observe that $$\left(1+\frac{x}{n}\right) ^n=1+x+\sum_{k=2}^n\frac{x^k}{k!}\left(1-\frac{1}{n}\right)\dots\left(1-\frac{k-1}{n}\right)=\sum_{k=0}^{n}a_k\frac{x^k}{k!}\tag{1}$$ where $$a_0=a_1=1,a_k=\left(1-\frac{1}{n}\right)\dots\left(1-\frac{k-1}{n}\right),k=2,\dots,n$$ Note that the coefficients $a_k$ are positive and do not exceed $1$. If $x>0$ then we can use the implication $$a_k\leq 1\implies a_kx^k\leq x^k\tag{2}$$ to bound the sum in $(1)$ with $\sum_{k=0}^n x^k/k! $. But if $x<0$ the implication $(2)$ does not hold for odd values of $k$ (rather the inequality gets reversed) and hence we can't find an upper bound.

Next I explain the argument used in reference 3 (my blog). Let $$F(x, n) =\left(1+\frac{x}{n}\right)^n,G(x,n)=\left(1-\frac{x}{n}\right)^{-n}\tag{3}$$ then we have $$F(-x, n) G(x, n) =1\tag{4}$$ Let $x>0$ and then we have already established (as in my blog or your question) that $F(x, n) $ is increasing (as function of $n$) and also bounded above and hence the limit $\lim_{n\to\infty} F(x, n) $ exists.

To handle negative values of $x$ I treat the expression $F(-x, n) $ with $x>0$. A better approach would have been to assume $x<0,x=-y,y>0$ and focus on $F(-y, n) $ but I have reused the symbol $x$ instead of inventing another symbol $y$. Thus I keep $x>0$ and handle both $F(x, n), F(-x, n) $.

Now as in reference 3 we have $x>0$ and $n>x$ so that we can apply general binomial theorem to write an infinite power series (in $x$) for $G(x, n) $ as $$G(x, n) =\sum_{k=0}^{\infty}b_k\frac{x^k}{k!}\tag{5}$$ where $$b_0=b_1=1,b_k=\left(1+\frac{1}{n}\right)\dots\left(1+\frac {k-1}{n}\right), k>1$$ Note that $b_k\geq 1$ and if $n$ increases then $b_k$ decreases and thus $G(x, n) $ is a decreasing sequence and bounded below by $\sum_{k=0}^{\infty}x^k/k!$ and hence bounded below by $1+x$. Thus $G(x, n) $ tends to a limit which is not less than $1+x$ (so that the limit is positive). Then $F(-x, n) =1/G(x,n)$ also tends to a positive limit.

Your own approach is smarter because it can show the increasing nature of $F(x, n) $ for all $x$ using Bernoulli inequality whereas I had to invent a $G(x, n) $ to deal with some cases.

Answer 2

• You application of Bernoulli's inequality is correct as long as $y_n=-\frac{x}{(n+x)(n+1)}\geq-1$. In reference (2) which you mention in your posting, Mark Viola assumes $x> -1$, in which case $y_n>-1$ for all $n\geq1$ and so, the sequence $f_n$ (in your notation) is monotone increasing for all $n\in\mathbb{N}$. However, he also points out that for any $x$, one can consider the first $n_0\in\mathbb{N}$ so that $n_0+x>0$. Then $y_n>-1$ for all $n\geq n_0$ and so, the sequence $f_n$ is monotone increasing for all for $n\geq n_0$. This shows that the limit exists for all $x\in\mathbb{R}$.

• An alternative approach shows that it is enough to consider the existence of the limit $e=\lim_n(1+\tfrac1n)^n$ to show that $\lim_n(1+\tfrac{a}{n})^n$ exists and equals $e^a$ for all $a\in\mathbb{R}$.

Indeed, from the existence of $e=\lim_n(1+\tfrac1n)^n$, we have that
$\lim_{x\rightarrow\infty}(1+\tfrac{1}{x})^x$ ($x$ ranges along positive reals) exists and has limit $e$. To see this, notice that $n_x:=\lfloor x\rfloor\leq x\leq \lfloor x\rfloor +1 = n_x+1$ hence $$\big(1+\tfrac{1}{n_x+1}\big)^{n_x}\leq\big(1+\tfrac1x\big)^{n_x}\leq\big(1+\tfrac1x\big)^x\leq\big(1+\tfrac{1}{n_x}\big)^x\leq\big(1+\tfrac{1}{n_x}\big)^{n_x+1} $$ As $n_x\xrightarrow{x\rightarrow\infty}\infty$, one gets that $$ \lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x+1}\big)^{n_x}=\lim_{n_x\rightarrow\infty}\frac{\big(1+\tfrac{1}{n_x+1}\big)^{n_x+1}}{1+\tfrac{1}{n_x+1}}=e $$ and $$ \lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x}\big)^{n_x+1}=\lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x}\big)^{n_x}\big(1+\tfrac{1}{n_x+1}\big)=e $$ Therefore $\lim_{x\rightarrow\infty}\big(1+\tfrac1x\big)^x=e$. As a consequence, for $a>0$, we have $$\lim_{n\rightarrow\infty}\big(1+\tfrac{a}{n}\big)^n=\lim_{n\rightarrow\infty}\left(\big(1+\tfrac{a}{n}\big)^{\tfrac{n}{a}}\right)^a=e^a $$ Similarly, for $x<0$, let $y=-x$ so that $y>0$. $$\big(1+\tfrac{1}{x}\big)^x=\big(1-\tfrac1y\big)^{-y}=\left(\frac{1}{1-\frac1y}\right)^y=\left(1+\tfrac{\tfrac1y}{1-\tfrac1y}\right)^y=\left(1+\tfrac{1}{y-1}\right)^{y-1}\big(1+\tfrac{1}{y-1}\big)$$ Thus, $$\lim_{x\rightarrow-\infty}\big(1+\tfrac1x\big)^x=\lim_{y\rightarrow\infty}\left(1+\tfrac{1}{y-1}\right)^{y-1}\big(1+\tfrac{1}{y-1}\big)=\frac1e$$ As a consequence, for $a>0$ $$\lim_{n\rightarrow\infty}\big(1-\tfrac{a}{n}\big)^n=\lim_{n\rightarrow\infty}\left(\big(1-\tfrac{a}{n}\big)^{-\tfrac{n}{a}}\right)^a=e^{-a}$$

• Notice that all these requires a definition of real powers that is a rigorous definition of $a^x$ for $a>0$ and $x\in\mathbb{R}$. This of course is done by starting with rational powers, show monotonicity and then use axioms such as that of the supremum or its equivalents. Other modern approaches are based on integration. One defines $\ln(x)=\int^x_1\frac{1}{t}\,dt$ for $x>0$ and shows that $\ln$ is a nice continuous, monotone increasing functions (furthermore, it is differentiable by the fundamental theorem of Calculus) taking values in all of $\mathbb{R}$, with the property that $\log(ab)=\log(a)+\log(b)$ for all $a,b>0$. Then, the exponential function is defined as the inverse of $\ln$. Theorems such as the inverse function theorem, would imply differentiability and the property that $\exp(a+b)=\exp(a)\exp(b)$. The number $e=\exp(1)$.