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This is a step on the way to proving $$ \lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n} = e^{z}.$$

I'm looking for an answer without 1) a summation or 2) a logarithmic function.

So far, I have that $$ \lim_{n \to \infty} |z_{n}| = \lim_{n \to \infty} \vert\left(1+\frac{z}{n}\right)^{n}\vert = \lim_{n \to \infty} \left(\sqrt{\left(1+\frac{x}{n}\right)^{2} +\left( \frac{y}{n}\right)^{2} }\right)^{n} = \lim_{n \to \infty}\left(\left( 1+\frac{x}{n}\right)^{2} + \left(\frac{y}{n} \right)^{2} \right)^{n/2} = \lim_{n \to \infty} \left( 1 + \frac{2x}{n} + \left(\frac{x}{n}\right)^{2} + \left( \frac{y}{n}\right)^{2}\right)^{n/2}. $$ Now, I was given the following hint, and I would like to use it, but I don't know how:

Hint: Argue that you can discard terms with $1/n^{2}$; for example, by showing that the limit $\displaystyle \lim_{n \to \infty} \frac{\left(\left( 1 + \frac{x}{n}\right)^{2} + \left(\frac{y}{n}\right)^{2}\right)^{n/2}}{\left(1 + \frac{2x}{n}\right)^{n/2}} = 1$.

How would I show this ($\epsilon$-definition?), and how would it help me here with what I've already done?


marked as duplicate by Alex M., Fabian, E. Joseph, Jack's wasted life, Qwerty Nov 2 '16 at 21:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ I'd try applying the binomial theorem to $(1+\frac{z}{n})^n$. $\endgroup$ – carmichael561 Feb 23 '16 at 4:22
  • $\begingroup$ @carmichael561, how would that help me here? $\endgroup$ – ALannister Feb 23 '16 at 4:43
  • $\begingroup$ Why the downvote for OP? It appeared to be a good question and without any votes, so I upvoted (and also provided an answer), but now that has been nullified by the downvote. $\endgroup$ – Paramanand Singh Feb 26 '16 at 4:17
  • $\begingroup$ @ParamanandSingh, because I threatened to downvote anybody who gave me an answer with summations or logarithms in my original post. $\endgroup$ – ALannister Feb 26 '16 at 18:21
  • $\begingroup$ @AlexM. this is not a duplicate. Please see what is written in boldface above. $\endgroup$ – ALannister Nov 3 '16 at 19:35

I think the best approach is the one which I found on MSE (will add link when I find it). It is based on the following general theorem.

Theorem: If $a_{n}$ is a sequence of real or complex terms such that $n(a_{n} - 1) \to 0$ as $n \to \infty$ then $a_{n}^{n} \to 1$ as $n \to \infty$.

The proof is done by writing $a_{n} = 1 + a_{n} - 1 = 1 + b_{n}$ (say) so that $nb_{n} \to 0$ as $n \to \infty$. Clearly we have \begin{align} a_{n}^{n} &= (1 + b_{n})^{n}\notag\\ &= 1 + nb_{n} + \dfrac{1 - \dfrac{1}{n}}{2!}(nb_{n})^{2} + \cdots\notag \end{align} and hence \begin{align} |a_{n}^{n} - 1| &\leq |nb_{n}| + \frac{1}{2!}|nb_{n}|^{2} + \cdots + \frac{1}{n!}|nb_{n}|^{n}\notag\\ &\leq|nb_{n}| + \frac{1}{2}|nb_{n}| + \frac{1}{2^{2}}|nb_{n}|^{2} + \cdots\notag\\ &= \dfrac{|nb_{n}|}{1 - \dfrac{|nb_{n}|}{2}}\notag\\ &\to 0\text{ as }n \to \infty\notag \end{align} Hence $a_{n}^{n} \to 1$ as $n \to \infty$.

It is now easy to show that if $z = x + iy$ then $$\left(1 + \frac{z}{n}\right)^{n} \to e^{x}(\cos y + i\sin y)$$ for all real $x, y$ where symbol $e^{x}$ is defined by the limit $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ I assume that you know the proof that the limit in $(1)$ exists for all real $x$ and this limit is non-zero.

I show that if $$f(z) = \lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n}\tag{2}$$ then $$f(z_{1} + z_{2}) = f(z_{1})f(z_{2})\tag{3}$$ for all $z_{1}, z_{2} \in \mathbb{C}$. Consider the sequence $$a_{n} = \dfrac{1 + \dfrac{z_{1} + z_{2}}{n}}{\left(1 + \dfrac{z_{1}}{n}\right)\left(1 + \dfrac{z_{2}}{n}\right)}\tag{4}$$ and then $$n(a_{n} - 1) = -\left(\dfrac{\dfrac{z_{1}z_{2}}{n}}{\left(1 + \dfrac{z_{1}}{n}\right)\left(1 + \dfrac{z_{2}}{n}\right)}\right)$$ so that $n(a_{n} - 1) \to 0$ as $n \to \infty$. Hence $a_{n}^{n} \to 1$ as $n \to \infty$. From definition of $a_{n}$ in equation $(4)$ it follows that $f(z_{1} + z_{2}) = f(z_{1})f(z_{2})$ provided that both the limits $f(z_{1}), f(z_{2})$ exist and are non-zero.

Next I show that the limit $f(z)$ defined by equation $(2)$ exists and is non-zero for all $z \in \mathbb{C}$. If $z$ is real then by assumption mentioned after equation $(1)$, the limit $f(z)$ exists and is non-zero. Next let $z = iy$ be totally imaginary. I will show that in this case $f(z) = \cos y + i\sin y$ and hence it is non-zero. Consider again the sequence $$c_{n} = \dfrac{1 + \dfrac{iy}{n}}{\cos\left(\dfrac{y}{n}\right) + i\sin\left(\dfrac{y}{n}\right)} = \cos\left(\frac{y}{n}\right) + \left(\frac{y}{n}\right)\sin\left(\frac{y}{n}\right) + i\left\{\left(\frac{y}{n}\right)\cos\left(\frac{y}{n}\right) -\sin\left(\frac{y}{n}\right)\right\}$$ and it is easy to see that $n(c_{n} - 1) \to 0$ as $n \to \infty$ and hence $c_{n}^{n} \to 1$ and from defining equation for $c_{n}$ we can see that denominator of $c_{n}^{n}$ is $(\cos y + i\sin y)$. It follows that $$(1 + iy/n)^{n} \to (\cos y + i\sin y)$$ Thus we have established that $f(z)$ exists and is non-zero in both the cases when $z$ is purely real and when $z$ is purely imaginary.

From equation $(3)$ it follows that if $z = x + iy$ then $f(z) = f(x)f(iy)$ and $f(z)$ is non-zero because $f(x), f(iy)$ are non-zero. We thus have $f(z) = e^{x}(\cos y + i\sin y)$.

It is easy to use the theorem mentioned in the beginning of my post to prove the result which OP is seeking in order to develop a proper theory of the limit of $(1 + z/n)^{n}$. Thus I show directly that if $z_{n} = (1 + z/n)^{n}$ then $|z_{n}|^{2} \to e^{2x}$ where $z = x + iy$ and hence $|z_{n}| \to e^{x}$. Clearly as mentioned by OP we have $$|z_{n}|^{2} = \left\{\left(1 + \frac{x}{n}\right)^{2} + \frac{y^{2}}{n^{2}}\right\}^{n}$$ Taking $$a_{n} = \dfrac{\left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}}{1 + \dfrac{2x}{n}}$$ we can see that $n(a_{n} - 1) \to 0$ and hence $a_{n}^{n} \to 1$ so that $|z_{n}|^{2} \to e^{2x}$.

OP has asked in the comments about choosing an appropriate sequence $a_{n}$ to make use of the theorem mentioned in my post. The basic idea is that the theorem is supposed to be used in situations where you want to find the limit of an expression whose exponent is $n$. Thus say you want to find limit of a sequence of type $c_{n}^{n}$. Moreover you intuitively know the answer as to what the limit will be. The idea is to find another sequence of type $d_{n}^{n}$ whose limit you already know and then set $a_{n} = c_{n}/d_{n}$.

Thus for example in OP's question there is a hint that the limit of $\left(\left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}\right)^{n/2}$ is related to that of the sequence $(1 + 2x/n)^{n/2}$. Squaring these sequences we get the exponents as $n$ and then it is a simple matter to figure out that $$c_{n} = \left(1 + \dfrac{x}{n}\right)^{2} + \dfrac{y^{2}}{n^{2}}, d_{n} = 1 + \frac{2x}{n}, a_{n} = \frac{c_{n}}{d_{n}}$$ A better case can be made for the limit $(1 + iy/n)^{n}$. We intuitively know that the limit should be $e^{iy} = \cos y + i\sin y$. Let's then try to find another expression whose $n^{\text{th}}$ power tends to $\cos y + i\sin y$. Here we are damn lucky because of De Moivre's Theorem $$\left(\cos\frac{y}{n} + i\sin\frac{y}{n}\right)^{n} = \cos y + i\sin y$$ and therefore we set $$c_{n} = 1 + \frac{iy}{n}, d_{n} = \cos\frac{y}{n} + i\sin\frac{y}{n}, a_{n} = \frac{c_{n}}{d_{n}}$$

  • $\begingroup$ interesting. I'll take a closer look at this in the morning, but it looks promising. $\endgroup$ – ALannister Feb 23 '16 at 5:04
  • 1
    $\begingroup$ @JessyCat: The approach assumes that you already know how to deal with $(1 + z/n)^{n}$ when $z$ is real and I must say the theorem mentioned in my answer is powerful enough to be used in many contexts. For example. It can be directly used to show that the modulus of $(1 + z/n)^{n}$ tends to $e^{\Re(z)}$ (this was your original question, I rather provided a full development of the theory of $(1 + z/n)^{n}$). $\endgroup$ – Paramanand Singh Feb 23 '16 at 5:09
  • $\begingroup$ could you please explain to me how to get $a_{n}$ from $(1+\frac{x}{n})^{2} + \frac{y^{2}}{n^{2}}$? One of the biggest problems I have had is figuring out where that expression for $a_{n}$ comes from. If you could write out all the intermediate steps to show where it came from, I would much appreciate it. $\endgroup$ – ALannister Feb 23 '16 at 21:53
  • $\begingroup$ @JessyCat: I have updated my answer. See last 2 paragraphs. $\endgroup$ – Paramanand Singh Feb 24 '16 at 3:38

Let $z=x+iy$. Then:

$$\left(1+\frac{x\pm iy}{n}\right)^n=\frac{\left(1+\frac{x}{n}\right)^n\left(1+\frac{\pm iy}{n+x}\right)^{n+x}}{\left(1+\frac{\pm iy}{n+x}\right)^x}\to\frac{e^xe^{\pm iy}}{(1+0)^x}=e^{x\pm iy}$$


$$z_n \to e^{x+iy}, \bar z_n\to e^{x-iy}\implies \vert z_n \vert ^2 \to e^{x+iy}e^{x-iy}=e^{2x}\implies \vert z_n \vert \to e^{x}=e^{\Re z}$$

  • $\begingroup$ Perhaps you could have said to take the principal values when the exponent is x or n+x . +1. $\endgroup$ – DanielWainfleet Feb 24 '16 at 17:04

Just show that

$$\left(1 + \frac{2x}{n} + \frac{D}{n^2}\right)^n = \sum_{k=0}^n \left(1+\frac{2x}{n}\right)^{n-k} C_{n}^k \frac{D^{k}}{n^{2k}} = (1 + \frac{2x}{n})^n + \sum_{k=1}^{n} (1 + \frac{2x}{n})^{n-k}C_n^k \frac{D^k}{n^{2k}}$$

the latter term take $n\to \infty$, there is a constant $M$ that $(1 + \frac{2x}{n})^{n-k} \le M$ for all $k$. thus we just need to show that

$$\sum_{k=1}^{n} C_n^k \frac{D^k}{n^{2k}}\to 0$$

$$\sum_{k=1}^{n} C_n^k \frac{D^k}{n^{2k}} = \sum_{k=1}^{n} \frac{n(n-1)\dots (n-k+1)}{k(k-1)\dots 1} \frac{D^k}{n^{2k}}$$


$$\frac{n}{k}\le \frac{n-1}{k-1}\le \dots \le \frac{n-k+1}{1}$$

we just have to show

$$\sum_{k=1}^{n}\left(n-k+1\right)^k\frac{D^k}{n^{2k}}\to 0$$


$$(n-k+1)\le n$$

$$\sum_{k=1}^{n}\left(n-k+1\right)^k\frac{D^k}{n^{2k}}\le \sum_{k=1}^{n} \frac{D^kn^k}{n^{2k}} = \sum_{k=1}^{n}\frac{D^k}{n^k} \le \sum_{k=1}^{\infty} \frac{D^k}{n^k} = \frac{D}{n-D}\to 0$$

  • $\begingroup$ I can't use this. I'm sorry. I asked for a solution with the hint I provided, and no sums. $\endgroup$ – ALannister Feb 23 '16 at 4:36
  • $\begingroup$ It is hard to see how you are going to define $e^z$ for complex $z$ without its power series. Perhaps you can try to show that the real function $\exp$ can be extended to an analytic function, but to do this without the use of power series seems like doing carpentry without a hammer. $\endgroup$ – DanielWainfleet Feb 23 '16 at 4:48
  • $\begingroup$ @user254665, there is a way to do it. I cannot assume that I know the power series expansion for $e$ here. What I think our prof wants us to do is somehow get rid of all terms in the limit with a $n^{2}$ in the , but I don't understand his hint on how to do that. Also, remember that I am trying to find what the modulus of $(1+\frac{z}{n})^{n}$ is, not $(1+\frac{z}{n})^{n}$ itself. So it's going to approach a real number. $\endgroup$ – ALannister Feb 23 '16 at 4:52
  • $\begingroup$ @user254665, this is just a first step. Part (b) of this question asks me to do the same thing for the arg. Then, finaly, in Part (c), I'm supposed to put it all together. $\endgroup$ – ALannister Feb 23 '16 at 4:57
  • $\begingroup$ Are you allowed to use $\lim_{n\to \infty}(1+x/n)^n=e^x$ for real $x$? $\endgroup$ – DanielWainfleet Feb 23 '16 at 5:17

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