# Prove that the limit definition of the exponential function implies its infinite series definition.

Here's the problem: Let $x$ be any real number. Show that $$\lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \sum_{n=0}^ \infty \frac{x^n}{n!}$$ I'm sure there are many ways of pulling this off, but there are 3 very important hints to complete the exercise in the desired manner:

1. Expand the left side as a finite sum using the Binomial Theorem. Call the summation variable $n$.
2. Now add into the finite sum extra terms which are $0$ for $n>m$, in order to make it look like an infinite series.
3. What happens to the limit on $m$ outside the series?

So far I was able to use Hint 1 to expand the left side: $$\lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \lim_{m \to \infty} \sum_{n=0}^m \binom {m}{n} \left( \frac{x}{m} \right)^n$$

No matter what I do with the binomial coefficients and factorials, I can't figure out what extra terms to add per Hint 2. Any suggestions?

• Show both limits converge. Take derivatives on both sides and evaluate at zero. You get the same thing in both. – user119256 Jan 13 '14 at 19:05
• This answer goes into detail on this question. – robjohn Jan 13 '14 at 19:58
• Perhaps this link might prove itself useful. – Lucian Jan 14 '14 at 3:45
• math.stackexchange.com/questions/1557074/… – Robert Wolfe Jul 26 '18 at 0:58

The answer from robjohn linked in his comment is a must visit. It deals with the definition of $e^{x}$ as $\lim_{n \to \infty}(1 + 1/n)^{nx}$. Another direct approach starting with $(1 + x/n)^{n}$ is given below.

Let $$F_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}$$ so that $\lim_{n \to \infty}F_{n}(x) = F(x)$ exist and $F(x)$ is given by the convergent power series $$F(x) = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots = \sum_{k = 0}^{\infty}\frac{x^{k}}{k!}$$ First we assume that $x > 0$. We have to establish that $\lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} = F(x)$

If $n > x$ then we have $$\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \frac{n(n - 1)}{2!}\left(\frac{x}{n}\right)^{2} + \cdots + \frac{n!}{n!}\left(\frac{x}{n}\right)^{n}$$ so that $$\left(1 + \frac{x}{n}\right)^{n} < F_{n}(x)$$

Next consider the expression $(1 - (x/n))^{-n}$. By the binomial theorem for negative exponents we have $$\left(1 - \frac{x}{n}\right)^{-n} = 1 + x + \frac{n(n + 1)}{2!}\left(\frac{x}{n}\right)^{2} + \cdots$$ so that $$\left(1 - \frac{x}{n}\right)^{-n} \geq F_{n}(x)$$ It follows that for $0 < x < n$ we have $$\left(1 + \frac{x}{n}\right)^{n} < F_{n}(x) \leq \left(1 - \frac{x}{n}\right)^{-n}\,\,\,\cdots (1)$$

Now note that $\lim_{n \to \infty}(1 + x/n)^{n} = G(x)$ exists and we have $(1 + (x/n))^{n} < G(x)$ for all $x > 0$ and all positive integers $n$. Therefore

\displaystyle \begin{aligned}\left(1 - \frac{x}{n}\right)^{-n} - \left(1 + \frac{x}{n}\right)^{n} &= \left(1 + \frac{x}{n}\right)^{n}\left\{\left(1 - \frac{x^{2}}{n^{2}}\right)^{-n} - 1\right\}\\ &< G(x)\left\{\left(1 - \frac{x^{2}}{n}\right)^{-1} - 1\right\} = \frac{x^{2}G(x)}{n - x^{2}}\end{aligned}

so that the expression $(1 - (x/n))^{-n} - (1 + (x/n))^{n}$ tends to $0$ as $n \to \infty$ (using squeeze theorem). It follows that $$\lim\limits_{n \to \infty}\left(1 - \dfrac{x}{n}\right)^{-n} = \lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} = G(x)\,\,\,\cdots (2)$$

Now applying squeeze theorem on equation $(1)$ and using equation $(2)$ above we get $\lim_{n \to \infty}F_{n}(x) = G(x)$ so that $F(x) = G(x)$ for all $x > 0$.

For $x = 0$ the result is trivial. For $x < 0$ we need to use the fact that $F(x + y) = F(x)F(y)$ (this is done by multiplication of series) and thus $F(-x) = 1/F(x)$. Similarly from equation $(2)$ it follows that $G(-x) = 1/G(x)$. Thus the relation $F(x) = G(x)$ holds for every real $x$. Noting the definitions of $F, G$ we see that $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots$$ for all $x \in \mathbb{R}$.

Update: While providing an answer to another question I came up with an alternative proof of $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = \lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^{-n}$$ which is worth mentioning here.

Let us set $$f(n) = \left(1 + \frac{x}{n}\right)^{n}, g(n) = \left(1 - \frac{x}{n}\right)^{-n}$$ and as in the earlier proof we only need to handle the case for $x > 0$. By the use of binomial theorem we can show that the sequence $f(n)$ increases and is bounded by $F(x)$ so that $f(n)$ tends to a limit as $n \to \infty$ and the limit is greater than or equal to $(1 + x)$.

Now we can see that if $0 < x^{2} < n$ then \begin{aligned}\frac{f(n)}{g(n)} &= \left(1 - \frac{x^{2}}{n^{2}}\right)^{n}\\ &= 1 - n\cdot\frac{x^{2}}{n^{2}} + \frac{n(n - 1)}{2!}\left(\frac{x^{2}}{n^{2}}\right)^{2} - \cdots + \\ &= 1 - \frac{x^{2}}{n} + \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} - \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\ &= 1 - \phi(n)\end{aligned} where $\phi(n)$ is a finite sum defined by $$\phi(n) = \frac{x^{2}}{n} - \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} - \cdots$$ We can estimate $\phi(n)$ as follows \begin{aligned}0 \leq |\phi(n)| &\leq \frac{x^{2}}{n} + \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\ &\leq \frac{x^{2}}{n} + \frac{1}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \frac{1}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\ &\leq \frac{x^{2}}{n} + \frac{1}{2}\left(\frac{x^{2}}{n}\right)^{2} + \frac{1}{2^{2}}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\ &= \frac{x^{2}}{n}\left\{1 + \frac{x^{2}}{2n} + \left(\frac{x^{2}}{2n}\right)^{2} + \cdots\right\}\\ &= \frac{x^{2}}{n}\cdot\dfrac{1 - \left(\dfrac{x^{2}}{2n}\right)^{n}}{1 - \dfrac{x^{2}}{2n}}\\ &\leq \frac{x^{2}}{n}\cdot\dfrac{1}{1 - \dfrac{x^{2}}{2n}}\to 0 \text{ as }n \to \infty\end{aligned} and then we see that $\phi(n) \to 0$ as $n \to \infty$. It follows that $f(n)/g(n) \to 1$ as $n \to \infty$ and therefore $\lim_{n \to \infty}f(n) = \lim_{n \to \infty}g(n)$.

$$\sum_{n=0}^m {m \choose n} \left( \frac{x}{m} \right)^n = \sum_{n=0}^\infty {m \choose n} \left( \frac{x}{m} \right)^n$$ since ${m \choose n} = 0$ for $n > m$.

Consider this. You are left with the following limit $$\frac{m!}{(m-n)! m^n}$$ Note that $$\frac{m!}{(m-n)!}$$ for $m\to \infty$ this term will tend to $m^n$ as it can be easily seen (well relatively easy, let me know if is not clear why). So our first formula tend to 1, so that your relationship is proved right.

• I'm not sure this works. Notice that the number of terms changes with $m$ as well, so I don't think it's enough to ask what each term does individually. – Ben Millwood Jan 13 '14 at 20:26
• As soon as I have time I will write a more detailed answer... – Umberto Jan 13 '14 at 20:30
• I don't think this works either... – Noldorin Jun 15 '16 at 15:11
• See my answer for some details on this. – Kagaratsch Jul 26 '18 at 0:58

To continue from where you left off: note that $\binom{m}{n} = 0$ if $n > m$, so the sum can be extended to an infinite sum: $$\lim_{m\to\infty} \left(1+\frac{x}{m}\right)^m = \lim_{m\to\infty} \sum_{n=0}^\infty \binom{m}{n} \left(\frac{x}{m}\right)^n.$$

Now, recall the Dominated Convergence Theorem for infinite sums:

Suppose $\sum_{n=0}^\infty b_n$ converges; for all $m,n$, $|a_{m,n}| \le b_n$; and for all $n$, $a_{m,n} \to c_n$ as $m\to\infty$. Then $\sum_{n=0}^\infty c_n$ converges, and $\sum_{n=0}^\infty a_{m,n} \to \sum_{n=0}^\infty c_n$ as $m\to\infty$.

To apply this, note that $$\binom{m}{n} \left(\frac{x}{m}\right)^n = \frac{x^n}{n!} \cdot \frac{m(m-1)\cdots(m-n+1)}{m^n} = \frac{x^n}{n!} \cdot \left(1-\frac{1}{m}\right)\left(1-\frac{2}{m}\right)\cdots\left(1-\frac{n-1}{m}\right).$$ Now, the absolute value of this term is at most equal to $\frac{|x|^n}{n!}$ (and in fact, it is less if $x\ne 0$ and $n>1$), and $\sum_{n=0}^\infty \frac{|x|^n}{n!}$ converges by the Ratio Test. Also, the limit of the term as $m\to\infty$ is equal to $\frac{x^n}{n!}$. Therefore, by the Dominated Convergence Theorem, $$\lim_{m\to\infty} \sum_{n=0}^\infty \binom{m}{n} \left(\frac{x}{m}\right)^n = \sum_{n=0}^\infty \frac{x^n}{n!}.$$

Recall that

$$\binom{m}{n}=\frac{m!}{n! (m-n)!}$$

By Stirling's formula you get for large $m$

$$m!\approx \sqrt{2\pi m}\left(\frac{m}{e}\right)^{m}\\ (m-n)!\approx \sqrt{2\pi(m-n)}\left(\frac{m-n}{e}\right)^{m-n}$$

Then note

$$\left(\frac{m-n}{e}\right)^{m-n}=e^{(m-n)\ln(m-n)-m+n}=e^{(m-n)\ln(m)+(m-n)\ln(1-\frac{n}{m})-m+n}$$

By Taylor series

$$f(x+\epsilon)=f(x)+f'(x)\epsilon+O(\epsilon^2)~~~\text{get}~~~\ln(1-\frac{n}{m})=\underbrace{\ln(1)}_{=0}-\frac{n}{m}+O(m^{-2})$$

and we proceed to

$$\left(\frac{m-n}{e}\right)^{m-n}\approx e^{(m-n)\ln(m)-(m-n)\frac{n}{m}-m+n}\approx e^{(m-n)\ln(m)-m}= \frac{m^{m-n}}{e^m}$$

where we neglected terms that certainly vanish in the limit $m\to\infty$.

Therefore, we conclude that

$$\frac{m!}{(m-n)!}= m^n+\text{subleading}$$

And all together

$$\sum_{n=0}^\infty \binom{m}{n} \left(\frac{x}{m}\right)^n= \sum_{n=0}^\infty \frac{m^n}{n!} \left(\frac{x}{m}\right)^n+\text{subleading}\to e^x$$