What is wrong with the following "proof" that $e=1$?

Question

Let's analyze this expression

$\lim_\limits{n\rightarrow\infty} (1+\frac{1}{n})^n$

It's the definition of $e$ which, as we know is not equal to $1$. So what is wrong with the following "logic":

As $\lim_\limits{n\rightarrow\infty}(a_{n}b_{n}) = \lim_\limits{n\rightarrow\infty}(a_{n})\times\lim_\limits{n\rightarrow\infty}(b_{n}) $ and $\lim_\limits{n\rightarrow\infty}(1+\frac{1}{n}) = 1$, we can say that $\lim_\limits{n\rightarrow\infty} (1+\frac{1}{n})^n=1^n$, which is equal to $1$.

I know something's wrong there, but the question is - what?

Answer 1

What you actually proved is that $$\lim_{k\to\infty} \left(\lim_{n\to\infty} 1+ \frac1n\right)^k = 1$$ Wich is correct, but the LHS is not equal to $e$.
The problem is most apparent when you end up with a $1^n$ (supposed to be a $\lim_{k\to\infty} 1^k$) and got rid of the limit expression $\lim_{n\to\infty}$.

Answer 2

You can apply your limit rule to two factors, and by induction to any fixed number of factors. But the number of factors in your limit is not fixed. You can't apply the limit as $n\to \infty$ to each of $n$ factors separately, because it is $n$ which is varying.

Answer 3

The limit $\lim_{n\to\infty} (1+\frac1n)^n$ works when both of the $n$s go towards infinity simultaneously. You can't split them up into two different variables that you move towards infinity one by one.

As a simpler example of the same effect, consider $$\lim_{n\to\infty} \frac1n\cdot n$$ This limit is obviously $1$, simply because $\frac1nn=1$ always. But it doesn't work to consider $\lim_{n\to\infty}\frac1n$ which is $0$ and then look at that other $n$ separately: $\lim_{n\to\infty}0\cdot n$ is $0$, not $1$.

Answer 4

There is no theorem which says that $$\lim_{n\to\infty} (a_n)^n=1$$ if $\lim_{n\to\infty} a_n=1$. In fact, this is false. The counterexaple is in front of you.

Answer 5

You're basically assuming things about limits that aren't true.

Let: $$a_{m,n}=\begin{cases}0&\text{if }m<n\\\frac12&\text{if }m=n\\1&\text{if }m>n\end{cases}$$ Then:

• $\displaystyle\lim_{m\to\infty}\left(\lim_{n\to\infty}a_{m,n}\right)=0$
• $\displaystyle\lim_{n\to\infty}a_{n,n}=\frac12$
• $\displaystyle\lim_{n\to\infty}\left(\lim_{m\to\infty}a_{m,n}\right)=1$

So even though it looks like those limits should be the same, they're all different. (I should mention that none of those parentheses are necessary, by the way.)

You're committing quite a similar fallacy with your argument.

Answer 6

You start of correctly by mentioning the product rule of limits. And further we do obviously have $\lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right) = 1$. The issue is in the next step. First of all you can't write $$\lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^{n} = 1^{n}$$ because the value of limit itself can't have the variable $n$. You might proceed like $$\lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^{n} = \lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)\cdot\left(1 + \dfrac{1}{n}\right)\cdots n\text{ times }$$ but then you can't apply the product rule to each factor because this rule is applicable only when number of factors is independent of limit variable.

Answer 7

For a start $1^\infty \neq 1$. Your proof is a counter example.

Also, $\lim (a_n b_n) = \lim a_n \times \lim b_n$ as long as both limits are finite and only for a finite (fixed) number of factors. In the case of $(1 + \frac{1}{n})^n = (1 + \frac{1}{n}) (1 + \frac{1}{n}) \cdots (1 + \frac{1}{n})$ [$n$ times] so the number of factors is not finite.