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Problem

Given that $\lim\limits_{n \to \infty}\left(1+\dfrac{1}{n} \right)^n = e$, show that $1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots=e$.

Proof

By binomial theorem, we have

\begin{align*}\left(1+\dfrac{1}{n}\right)^n&=\sum_{k=0}^{k=n}\binom{n}{k}1^k \left(\frac{1}{n}\right)^{n-k}\\ &=\sum_{k=0}^{k=n}\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right). \end{align*}

Thus, on one hand, $$\left(1+\dfrac{1}{n}\right)^n \leq \sum_{k=0}^{k=n}\frac{1}{k!}.\tag1$$

Take the limits as $n \to \infty$ on the both sides of $(1)$. We have $$e=\lim_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n\leq \varliminf_{n \to \infty} \sum_{k=0}^{k=n}\frac{1}{k!}.\tag2$$

On the other hand, take a positive integer $m$ such that $m<n$ and fix it. We have

$$\left(1+\dfrac{1}{n}\right)^n \geq \sum_{k=0}^{k=m}\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).\tag3$$

Likewise,take the limits as $n \to \infty$ on the both sides of $(3)$. We have

$$e=\lim_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n\geq \sum_{k=0}^{k=m}\frac{1}{k!}.\tag4$$

Take the limits as $m \to \infty$ on the both sides of $(4)$. We have

$$e \geq \varlimsup_{m \to \infty}\sum_{k=0}^{k=m}\frac{1}{k!}.\tag 5$$

Combine $(2)$ and $(5)$. It follows that $$\lim_{n \to \infty}\sum_{k=0}^{k=n}\frac{1}{k!}=e.$$

Please correct me If I'm faulty. Hope to see other solutions.

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Another more elegant proof

At the beginning, let's recall an unusual fact, named Tannery's limit theorem, which states that

Let $S(n)=\sum\limits_{k=0}^{n}a_k(n)$. If the following conditions are all satisfied:

  • For any $k$, $\lim\limits_{n \to \infty}a_k(n)=a_k$;
  • for any $k$,there exists a $M_k>0$ independent of $n$ such that $|a_k(n)|\leq M_k$;
  • $\sum\limits_{k=0}^{\infty}M_k<\infty$,

then $$\lim\limits_{n \to\infty}S(n)=\sum\limits_{k=1}^{\infty}a_k.$$

As for the present problem, we may denote $$S(n)=\left(1+\frac{1}{n}\right)^n,$$ and$$a_k(n)=\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).$$ Thus, we may verify that:

  • For any $k$, $\lim\limits_{n \to \infty}a_k(n)=\dfrac{1}{k!}=a_k$;
  • for any $k$,$|a_k(n)|\leq \dfrac{1}{k!}=M_k$;
  • $\sum\limits_{k=0}^{\infty}\dfrac{1}{k!}=\sum\limits_{k=0}^{\infty}M_k<2+\sum\limits_{k=2}^{\infty}\dfrac{1}{2^{k-1}}=3<\infty$.

It follows that $$e=\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=\lim\limits_{n \to \infty}S(n)=\sum\limits_{k=0}^{\infty}\dfrac{1}{k!}=\sum\limits_{k=0}^{\infty}a_k.$$

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  • $\begingroup$ Why downvote? Counter criticism? $\endgroup$ – xbh Aug 13 '18 at 16:35
  • $\begingroup$ Feels like you could refer to $M_k$ as $\varepsilon_k$ to show what's happening. But it's the same difference. $\endgroup$ – Jam Aug 13 '18 at 21:16
  • $\begingroup$ Tannery's M-Test? $\endgroup$ – BCLC Aug 15 '18 at 10:55
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\begin{eqnarray*} \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n} &=& \lim_{n\to \infty} \sum_{k=0}^{n}{\binom{n}{k} \left(\frac{1}{n}\right)^{k}} \\ &=& \lim_{n\to \infty} \sum_{k=0}^{n}\frac{1}{k!}{\prod_{m=0}^{k-1}{\left(1-\frac{m}{n}\right)} } \\ &=& \sum_{k=0}^{\infty}{\frac{1}{k!}\prod_{m=0}^{k-1}{ \lim_{n\to \infty} \left(1-\frac{m}{n}\right)} } \end{eqnarray*}

The key thing is that the limit of $\prod_{m=0}^{k-1}{\left(1-\frac{m}{n}\right)}$ as $n$ grows without bound becomes $1$.

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  • 1
    $\begingroup$ Wrong! you can't like that. It's conditional to take the limit into it. $\endgroup$ – mengdie1982 Aug 13 '18 at 16:14
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    $\begingroup$ Otherwise, you may do like this $\lim\limits_{n\to \infty} \left(1+\frac{1}{n}\right)^{n}=\lim\limits_{n\to \infty} (1+\frac{1}{n})\lim\limits_{n\to \infty} (1+\frac{1}{n})\cdots=1\cdot 1\cdot 1\cdots=1.$ Isn't this more direct? $\endgroup$ – mengdie1982 Aug 13 '18 at 16:17
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    $\begingroup$ You can take the limit inside sum, if it is bounded. Product is different. Not sure, you really wanted to make a statement here, by invoking an infinite product (which will go unbounded, even if the underlying term is bounded). $\endgroup$ – NivPai Aug 13 '18 at 16:25
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The idea of this approach is to define exponential function $\exp x$ by series, and then prove that $\exp 1 = e$. However, we use quite a few of standard properties of $\exp$ and $\ln$, so potential caveat is that some of these properties might not be easily proven by using such definition of $\exp x$, but I was not able identify any such weak part (required properties are listed below, scrutiny is welcome).

Definition 1. $e:=\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n$.

Definition 2. $\exp(x):=\sum_{n\geq0}\frac{x^n}{n!}$.

Some useful standard results:

  • $\exp x$ is strictly monotone (increasing) on $\mathbb{R}$,
  • $\exp 0 = 1$ (follows from definition)

Then we are justified in following definition:

Definition 3. Let $\ln x$ be an inverse of $\exp(x)$, i.e. $\ln(\exp(x))=\exp(\ln(x))=x$.

For the proof we will require also these properties:

  • $\ln x$ is continuous,
  • $\ln x$ is strictly increasing,
  • $\ln x^n=n \ln x$.
  • $\frac{d}{dx} \ln x=\frac{1}{x}$. (follows from derivative of inverse function and from $(\exp x)'=\exp x$, which can be shown from definition)
  • $\ln 1 = 0$ (follows from $\exp 0 = 1$ and inverse)

Let's first show that $\ln x=1$ has unique solution and this solution is $e$ (proof inspired by https://proofwiki.org/wiki/Euler%27s_Number:_Limit_of_Sequence_implies_Base_of_Logarithm). Indeed we have \begin{align} \ln e &= \ln\left(\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n\right)\\ &= \lim_{n \to \infty}\left(\ln \left(1+\frac{1}{n}\right)^n\right)\tag{1}\\ &= \lim_{n \to \infty}\frac{\ln \left(1+\frac{1}{n}\right)}{1/n}\\ &= \lim_{h \to 0}\frac{\ln \left(1+h\right)-\ln 1}{h}\\ &= \frac{d}{dx}\ln x|_{x=1}\tag{2}\\ &= \frac{1}{x}|_{x=1}\\ &= 1\\ \end{align} where in $(1)$ we used that $\ln x$ is continuous, and in $(2)$ we have used the definition of derivative. Since the $\ln x$ is injective, $e$ is the unique solution.

Now it follows $$e = \exp (\ln e) = \exp(1) =\sum_{n\geq0}\frac{1}{n!},$$ which is what we wanted to prove.

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