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


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.


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

  • $\begingroup$ 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

\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 $1$.

  • 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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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