Proof: $\lim \limits_{n\to\infty}(1+\frac{z}{n})^n = \exp(z)$

Question

I define $\exp: \mathbb C \to \mathbb C$ as $z \mapsto \sum \limits_ {k=0}^{\infty}\frac{z^k}{k!}$. I would like to show that $\lim \limits_{n\to\infty}(1+\frac{z}{n})^n = \exp(z)$. I have a proof for the case $z \in \mathbb R$, but the proof assumes that $\lim \limits_ {n\to\infty}(1+\frac{z}{n})^n$ exists, which is easy to see if $z \in \mathbb R$, but not that easy (for me) if $z \in \mathbb C$.

Would be good to have a proof which does not use derivatives or integrals.

Answer 1

I give another approach based on same technique as mentioned in this answer, but a lot simpler. Note that using the series definition of $\exp(z)$ you can prove that $$\exp(z + w) = \exp(z)\exp(w)$$ and therefore $\exp(z)\exp(-z) = 1$ so that $\exp(z) \neq 0$ for all $z \in \mathbb{C}$.

Now consider the sequence $$a_{n} = \dfrac{1 + \dfrac{z}{n}}{\exp\left(\dfrac{z}{n}\right)} = \left(1 + \frac{z}{n}\right)\exp(-z/n) = 1 - \frac{z^{2}}{n^{2}} + \dots$$ where $\dots$ represent terms with higher powers of $z/n$ so that we can write $$a_{n} = 1 - \frac{z^{2}}{n^{2}} + o(1/n^{2})$$ and therefore $n(a_{n} - 1) \to 0$ as $n \to \infty$. It follows from the theorem mentioned in the linked answer that $a_{n}^{n} \to 1$ and hence $$\lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n} = \exp(z)$$

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The simplicity of this approach is because of the series representation of $\exp(z)$. In the linked answer the series for $\exp(z)$ is not used and instead I prove that if $z = x + iy$ then $$\lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n} = e^{x}(\cos y + i\sin y)$$ where $e^{x}$ is defined by $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ Combining the current answer and the linked answer we see that for $z = x + iy$ we have $$\exp(z) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!} = e^{x}(\cos y + i\sin y)$$ and hence putting $x = 0$ and comparing real and imaginary parts we can get the series expansions for $\sin y$ and $\cos y$ valid for all real $y$.

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It is interesting to note that I had the linked answer available with me for almost a year and yet it took your question to use the same technique to derive power series for circular functions. Things like these never cease to amaze me!

Answer 2

Set, for $n\geq 1$, $$ (1+\dfrac{z}{n})^n:=\sum_{k=0}^\infty a(n,k)z^k $$ (finite support, it is a polynomial). It can be checked easily (I can elaborate on request) that $a(n,k)$ is increasing in $n$ and then for all $j\geq 2$, one has

$$|a(j,k)-a(j-1,k)|=a(j,k)-a(j-1,k)$$

Then, for all fixed $z\in \mathbb{C}$, the family $\Big((a(j,k)-a(j-1,k))z^k\Big)_{j\geq 2\atop k\geq 0}$, is absolutely summable as

$$ \sum_{k\geq 0}|z^k|\sum_{j\geq 2}(a(j,k)-a(j-1,k))=\sum_{k\geq 0}|z|^k \Big(\frac{1}{k!}-a(1,k)\Big)<+\infty\ . $$

Now, as this family is absolutely (and then commutatively) summable, one has, summing it by columns
$$ \sum_{k\geq 0}z^k\Big(\frac{1}{k!}-a(1,k)\Big)=e^z-(1+z) $$ and, on the other hand, summing it by rows, $$ \lim_{N\to +\infty}\sum_{j=2}^N\Big(\sum_{k\geq 0}z^k(a(j,k)-a(j-1,k))\Big)=\lim_{N\to +\infty}\Big(\Big(1+\frac{z}{N}\Big)^N\Big)-(1+z) $$ Thanks to @ParamanandSingh for having pointed the rôle of monotonicity (this is the golden mine of interactions).

Note This result (and proof) holds for $$ \lim_{n\to +\infty} \Big(1+\frac{A}{n}\Big)^n=\exp(A) $$ where $A$ is a element in a complete associative commutative normed algebra (over $\mathbb{R}$ or even $\mathbb{Q}$ as only divisibility is used).

Answer 3

Note that Taylor's Inequality is valid for complex functions.

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Let $$f:\;z\mapsto\lim_{n\to\infty}(1+{z\over n})^n.$$ One can prove that:

1. $f$ is well-defined;
2. $f$ is differentiable and $f'=f$ (thus $f \in \mathscr C^{\infty}$).

Then, choosing $M=3^r$ for example, one has $$\forall r>0,\;\exists M>0,\;\forall z\in D(0,r),\;\forall n\in\Bbb N,\;|f^{(n)}(z)|\le M.$$ Hence, (using Taylor's Inequality), $f$ can be expressed by its Taylor series in $\Bbb C$, which is to say, $$\forall z\in \Bbb C,\;f(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}.$$

Answer 4

Background Information:

I am writing this to clarify the proof in the first part of @Paramanand Singh's answer and add my own understanding to it.

When I was reading his answer, I think his idea is very interesting, but for some points, for example the claim that $a_n = 1 - \frac{z^2}{n^2} + o(1/n^2)$, they don't seem obvious to me, and I decide to restart the proof given his idea.

Below is my attempt based upon @Paramanand Singh's answer.

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My Attempt:

First of all, if we define the exponential function and the constant $e$ as

Definition 1. $$ \forall z \in \mathbb{C}, \exp(z) := \sum_{n=0}^{\infty} \frac{z^n}{n!}$$ by ratio test it is easy to show that this series indeed converges for all $z \in \mathbb{C}$.

Definition 2. $$ e := \exp(1) = \sum_{n=0}^{\infty} \frac{1}{n!}$$

Then we have the following important theorems and lemmas:

Theorem 1. (Euler's formula) $$\forall x \in \mathbb{R}, e^{ix} = \cos x + i \sin x$$

Theorem 2. (De Moivre's formula) $$\forall x \in \mathbb{R}, \forall n \in \mathbb{Z}, (\cos x + i \sin x)^n = \cos nx + i \sin nx$$

Lemma 1. (multiplication rule)

1. $\forall z,w \in \mathbb{C}, \exp(z+w) = \exp(z) \exp(w)$. Proof is easy by Cauchy Product.
2. $\forall z \in \mathbb{C}, 1 = \exp(z) \exp(-z)$.

Lemma 2. If $\displaystyle \lim_{n\to \infty} n(a_n-1)=0$, then $\displaystyle \lim_{n\to \infty} a_n^n=1$. For proof, see this answer.

Lemma 3. (real exponential function as a power of $e$) $$\forall x \in \mathbb{R}, \exp(x) = e^x$$ For sketch of proof, see this answer.

Then, by Lemma 2, fix some $z \in \mathbb{C}$ and still consider the sequence

$$a_n = \frac{ 1 + \frac{z}{n}}{\exp(\frac{z}{n})}$$

and we have

$$\begin{aligned} \frac{ 1 + \frac{z}{n}}{\exp\left(\frac{z}{n}\right)} &= \left(1 + \frac{z}{n} \right) \left(\exp\left(\frac{z}{n}\right)\right)^{-1} \\ &= \left(1 + \frac{z}{n} \right) \exp\left(-\frac{z}{n}\right) \\ &= \left(1 + \frac{z}{n} \right) \sum_{k=0}^{\infty} \frac{1}{k!} \left(-\frac{z}{n}\right)^k \\ &= \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left(\frac{z}{n}\right)^k + \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left(\frac{z}{n}\right)^{k+1} \\ &= 1 +\sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \left(\frac{z}{n}\right)^k + \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left(\frac{z}{n}\right)^{k+1} \end{aligned}$$

Therefore, we have

$$\begin{aligned} 0 \leq \left| n\left( \frac{ 1 + \frac{z}{n}}{\exp\left(\frac{z}{n}\right)} - 1 \right)\right| &= \left|\sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \frac{z^k}{n^{k-1}} + \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \frac{z^{k+1}}{n^k} \right| \\ &\leq |z|\sum_{k=1}^{\infty} \left(\frac{|z|}{n}\right)^{k-1} + |z|\sum_{k=0}^{\infty} \left(\frac{|z|}{n}\right)^{k} \\ &= 2|z|\sum_{k=0}^{\infty} \left(\frac{|z|}{n}\right)^k \end{aligned}$$

Then, since $z$ is fixed, as $n \to \infty$, we have $\frac{|z|}{n} \to 0 < 1$. And this implies

$$\begin{aligned} 0 \leq \lim_{n \to \infty}\left| n\left(\frac{ 1 + \frac{z}{n}}{\exp\left(\frac{z}{n}\right)} - 1 \right)\right| &\leq \lim_{n \to \infty} 2|z|\sum_{k=0}^{\infty} \left(\frac{|z|}{n}\right)^k \\ &= 2|z|\lim_{n \to \infty} \lim_{m \to \infty}\sum_{k=0}^{m} \left(\frac{|z|}{n}\right)^k \\ &= 2|z|\lim_{n \to \infty} \lim_{m \to \infty} \frac{1 - (\frac{|z|}{n})^{m+1}}{1 - \frac{|z|}{n}} \\ &= 2\lim_{n \to \infty} \frac{\frac{|z|}{n}}{1 - \frac{|z|}{n}} \\ &= 2 \times \frac{0}{1 - 0} \\ &= 0 \end{aligned}$$

So by squeeze theorem, we have

$$\lim_{n \to \infty}\left| n\left(\frac{ 1 + \frac{z}{n}}{\exp\left(\frac{z}{n}\right)} - 1 \right)\right| = 0$$

and this implies

$$\lim_{n \to \infty} n\left(\frac{ 1 + \frac{z}{n}}{\exp\left(\frac{z}{n}\right)} - 1 \right) = 0$$

Then, by Euler's formula and De Moivre's formula and Lemma 1 and Lemma 3, we have

$$\begin{aligned} \left(\exp \left(\frac{z}{n}\right)\right)^n &= \left(\exp \left(\frac{x}{n} + i\frac{y}{n}\right)\right)^n\\ &= \left(\exp\left(\frac{x}{n}\right)\exp\left(i\frac{y}{n}\right)\right)^n \\ &= \left(\exp\left(\frac{x}{n}\right)\right)^n \left(\exp\left(i\frac{y}{n}\right)\right)^n\\ &= \left(e^{\frac{x}{n}}\right)^n \left( \cos \frac{y}{n} + i \sin \frac{y}{n} \right)^n \\ &= e^x (\cos y + i \sin y)\\ &= \exp(x) \exp(iy)\\ &= \exp(x + iy) \\ &= \exp(z) \end{aligned}$$

So we have

$$\lim_{n \to \infty} \left(\frac{ 1 + \frac{z}{n}}{\exp\left(\frac{z}{n}\right)} \right)^n = \lim_{n \to \infty} \frac{ (1 + \frac{z}{n})^n}{(\exp(\frac{z}{n}))^n} = \frac{\lim_{n \to \infty} (1 + \frac{z}{n})^n}{\exp(z)} = 1$$

and this implies

$$ \lim_{n \rightarrow \infty}\left(1+\frac{z}{n}\right)^n = \exp(z) $$

And here we finish the proof. $\square$

Answer 5

By the Binomial Theorem:

$(1+\frac{z}{n})^n= \sum _{k=0}^n \binom{n}{k}\frac{z^k}{n^k}= \sum_{k=0}^n \frac{n!z^k}{k!(n-k)!n^k}$

The result holds if we prove $\lim_{n \to \infty} \frac{n!}{(n-k)!n^k}=1$

$\frac{n!}{(n-k)!n^k}=\frac{n(n-1)(n-2)...(n-k+1)}{n^k}=1(1-1/n)(1-2/n)(1-3/n)...(1- (k-1)/n)$

The product will be $1$ plus multiples of powers of $1/n$, so the limit is $1$.