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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.

zesy
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    See this answer http://math.stackexchange.com/a/1668179/72031 You need to know that the infinite series in question say $f(z) $ satisfies $f(z+w) =f(z) f(w) $ and $f(iy) =\cos y+i\sin y$. – Paramanand Singh Feb 16 '17 at 10:35
  • @ParamanandSingh. Your answer in the link is beautiful indeed. Cheers – Claude Leibovici Feb 16 '17 at 10:38
  • @ClaudeLeibovici: glad to know that you liked it! – Paramanand Singh Feb 16 '17 at 10:41
  • @ClaudeLeibovici: I added another answer based on the same technique. – Paramanand Singh Feb 17 '17 at 16:51
  • @DuchampGérardH.E.: for $z=1$ the result you are thinking can be derived using monotone convergence theorem. And it appears with suitable tweaking it can be extended to other real and complex values of $z$. – Paramanand Singh Feb 17 '17 at 17:19
  • "but the proof assumes that limn→∞(1+z/n)n exists, which is easy to see if z∈R, but not that easy (for me) if z∈C" $|(1 + z/n)^n - k| = |1 + ...a_k(z/n)^k + ..... + (z/n)^n -k|\le |1 + ...a_k(|z|/n)^k + ..... + (|z|/n)^n -k| =|(1+ |z|/n)^n - k|$. So if it converges for positive R it converges for C. – fleablood Feb 17 '17 at 17:25
  • @ParamanandSingh [And it appears with suitable tweaking]---> yes, it suffices to consider $\frac{z}{|z|}$. I'll fix the details for myself. Thank you for monotone pointing. – Duchamp Gérard H. E. Feb 17 '17 at 21:49
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    You may find my answer here helpful: http://math.stackexchange.com/questions/374747/if-z-n-to-z-then-1z-n-nn-to-ez/374765#374765 It uses the series expansion, and the property $\exp(z+w)=\exp(z)\exp(w)$ to prove a more general result, on the complex plane. –  Feb 18 '17 at 15:04
  • @Byron Schmuland your linked answer is indeed very helpful, I even managed to generalize is to the case $\lim_{n \to \infty}(1+\frac{z_n}{n})^n=\exp(\lim_{n \to \infty}z_n)$ for a converging $z_n$. – zesy Mar 09 '17 at 18:29
  • @SergeyZykov Glad to help out! –  Mar 09 '17 at 18:37

3 Answers3

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

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 $y$.

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!

Community
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Paramanand Singh
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  • Could you please elaborate on this equality $\left(1 + \frac{z}{n}\right)\exp(-z/n) = 1 - \frac{z^{2}}{n^{2}} + \dots$? – zesy Mar 09 '17 at 18:23
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    @SergeyZykov: use the power series for $\exp(-z/n) $ and multiply it with $(1+z/n)$ and you will get desired result. – Paramanand Singh Mar 10 '17 at 05:06
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Note that Taylor's Inequality is valid for complex functions.

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, $$\forall z\in \Bbb C,\;f(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}.$$

Aforest
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  • I am sorry but I need to downvote the answer as I explicitly asked for a proof without derivatives. – zesy Feb 16 '17 at 17:37
  • @SergeyZykov OK. But I prefer to say that the main idea is to prove that $|R_n(x)|\to0$. One can do it manually even without using Taylor's Inequality. And I suppose my proof involves Taylor expansion rather than derivatives. – Aforest Feb 17 '17 at 05:19
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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 $$ \sum_{k\geq 0}|z^k|\sum_{j\geq 2}(a(j,k)-a(j-1,k))=\sum_{k\geq 0}|z|^k(\frac{1}{k!}-a(1,k))<+\infty\ . $$ Now, as this family is absolutely (and then commutatively) summable, one has, summing it by columns
$$ \sum_{k\geq 0}z^k(\frac{1}{k!}-a(1,k))=e^z-(1+z) $$ and, on the other hand, summing it by rows, $$ lim_{N\to +\infty}\sum_{j=2}^N\Big(\sum_{k\geq 0}z^k(a(j,k)-a(j-1,k))\Big)=lim_{N\to +\infty}\Big((1+\frac{z}{N})^N\Big)-(1+z) $$ Thanks to @ParamanandSingh for having pointed the rôle of monotonicity (this is the golden mine of interaction).

Note This result (and proof) holds for $$ lim_{n\to +\infty} (1+\frac{A}{n})^n=exp(A) $$ where $A$ is a element in a complete associative commutative normed $\mathbb{R}$-algebra.

Duchamp Gérard H. E.
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