The definition of e by limits of $(1+1/n)^n$ through series expansion

Question

I think the problem I have is due to not being knowledgeable about limits.

If I use binomial expansion to expand $(1+1/n)^n$ to $1 + \frac{n!}{(n-k)!k!}*(1/n)^k + ...$, I can imagine replacing $n$ with $\infty$, so now I have:

$$ 1 + \frac{\infty!}{(\infty-k)!k!}*(1/\infty)^k + ... $$

Clearly by whatever method I don't understand, this reduces to the usually mentioned $1 + 1/k! + ...$

I am struggling to see how the expression is simplified when I can't see any obvious correspondence between the infinity numerators and denominators.

Answer 1

Most of the arguments given in other answers have a curious fallacy. In the binomial expansion of $(1+1/n)^{n}$ the number of terms as well as each term is dependent on $n$ hence taking limits term by term is not justified. A proper proof requires more analysis. I have presented this proof in detailed manner in my blog post.

Update: Upon OP's request (see comments below) I have provided the full proof below. Its a variation on the proof given in blog. But before that a few remarks regarding the fallacy of other answers are in order. Taking limits term by term is valid in two general contexts:

1) When the number of terms is finite and independent of the limit variable $n$.

2) When the number of terms is infinite (case of infinite series, uniform convergence)

The current example is an expression of type $$f(n) = \sum_{k = 0}^{n}g(n, k)$$ so that each term $g(n, k)$ depends on $n$ and total numbers of terms also depends on $n$. Let's have a counter example. Let $g(n, k) = 1/n$ for all $k, n$. And consider $$f(n) = \sum_{k = 1}^{n}g(n, k) = \sum_{k = 1}^{n}\frac{1}{n} = 1$$ If we take limits term by term we get $\lim_{n \to \infty}g(n, k) = 0$ so that term by term limit gives the infinite series $$0 + 0 + 0 + \cdots = 0$$ But since the function $f(n) = 1$ so the limit should have been $1$.

Now I come to the proof of the relation $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\tag{1}$$ Let us put $$f(n) = \left(1 + \frac{1}{n}\right)^{n}, g(n) = \left(1 - \frac{1}{n}\right)^{-n}, E(n) = \sum_{k = 0}^{n}\frac{1}{k!}$$ Note that the series $\sum (1/k!)$ is convergent so that $\lim_{n \to \infty}E(n)$ exists. For $n > 1$ I show that $$f(n) \leq E(n) \leq g(n)\tag{2}$$ The first inequality $f(n)\leq E(n)$ is obvious by the binomial expansion of $f(n)$ as each term in binomial expansion of $f(n)$ is less than or equal to corresponding term in $E(n)$. Since $n > 1$ so that $1/n < 1$ we can therefore apply binomial theorem for general exponents to get $$g(n) = 1 + n\cdot\frac{1}{n} + \frac{n(n + 1)}{2!}\cdot\frac{1}{n^{2}} + \cdots$$ which is an infinite series with positive terms. Now each term in above series is greater than or equal to each corresponding terms in $E(n)$. Moreover terms in $g(n)$ are infinite in number whereas $E(n)$ has a finite number of terms. Thus we have $E(n) \leq g(n)$ for all $n > 1$. So the inequalities mentioned in $(2)$ are established.

Again we know that $\lim_{n \to \infty}f(n) = e$ (I won't discuss the existence of this limit as it is a given assumption in the question, but its existence does require a proof). I now show that $$\lim_{n \to \infty}\frac{f(n)}{g(n)} = 1\tag{3}$$ so that $\lim_{n \to \infty}g(n)$ also exists and is equal to $e$. Once this is done we can apply squeeze theorem on $(2)$ and get $\lim_{n \to \infty}E(n) = e$ or what is the same as equation $(1)$. It thus remains only to show that equation $(3)$ holds.

Clearly we have $$\frac{f(n)}{g(n)} = \left(1 - \frac{1}{n^{2}}\right)^{n} = c(n)\text{ (say)}$$ and since $$0 < 1 - \frac{1}{n^{2}} < 1$$ it follows that $$1 - n\cdot\frac{1}{n^{2}} \leq c(n) \leq 1 - \frac{1}{n^{2}}$$ (note that first inequality follows from $(1 + x)^{n} \geq 1 + nx$ for $n > 1$ and $x > -1$). Applying squeeze theorem on the above inequalities we get $\lim_{n \to \infty}f(n)/g(n) = \lim_{n \to \infty}c(n) = 1$ and thus equation $(3)$ is verified. This completes the proof of $(1)$.

Answer 2

The Binomial Expansion is given by $$\begin{eqnarray*} \left(1+\frac{1}{n}\right)^{\!n} &=& 1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^{\!2}+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^{\!3}+\cdots \\ \\ &=& 1+1+\frac{n(n-1)}{n^2}\cdot\frac{1}{2!}+\frac{n(n-1)(n-2)}{n^3}\cdot \frac{1}{3!}+\cdots \\ \\ \end{eqnarray*}$$ The key thing to notice is that each or these coefficients in $n$ tend to $1$ as $n \to \infty$: $$\begin{eqnarray*} \lim_{n \to \infty} \frac{n(n-1)}{n^2} &=& \lim_{n \to \infty} 1-\frac{1}{n} &=& 1 \\ \\ \lim_{n \to \infty} \frac{n(n-1)(n-2)}{n^3} &=& \lim_{n \to \infty} \left(1-\frac{1}{n}\right)\!\left(1-\frac{2}{n}\right) &=& 1 \end{eqnarray*}$$ It follows that $$\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{\!n} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots = \mathrm{e}$$

Answer 3

I understood nothing of what you have written, but: $$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n \frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k=\sum_{k=0}^n \frac{1}{k!}\prod_{i=n-k+1}^n \frac{i}{n}\leq e$$ Because: $\frac{n!}{n-k!}=\prod_{i=n-k+1}^n i$ are exactly $k$ terms, so I can bring in the $\frac{1}{n^k}$.

If we take the limit with $n\to\infty$ then $$\prod_{i=n-k+1}^n \frac{i}{n}=\prod_{i=1}^k \frac{i+n-k}{n}\to 1$$

Edit: $$a_k(n)=\frac{1}{k!}\prod_{i=n-k+1}^n \frac{i}{n}\qquad k\leq n$$ $$a_k(n)=0\qquad\qquad\qquad\quad\,\, k>n$$ $a_k(n)$ is an increasing sequence of elements in $l^1$ i.e. $a_k(n)\leq a_k(n+1)$ for all $k$ and $n$, then for the monotone convergence theorem we can conclude that: $$\left(1+\frac{1}{n}\right)^n\to\sum_{k=0}^\infty \frac{1}{k!} \quad .$$

Have fun.

Answer 4

Note that $$\frac{n!}{(n-k)!n^{k}}=\frac{n(n-1)(n-2)....(n-k+1)}{\underbrace{n\cdot n \cdots n}_{\text{$k$ times}}}$$

So, as $n$ tends to infinity, it becomes $1$. I think you can figure out the rest.