<s>[INST] Why does a boundedness proof of $\exp(x)$ depend on the sign of $x$? 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:

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.

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:

*

*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?


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


*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.
 [/INST] 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.
</s>