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To be found: $$\lim \left(1+\frac{2}{n}\right)^n$$

Presuppose $~~\lim \left(1+\frac{1}{n}\right)^n=e~~$ is already shown.

Expanding the first equation: $$\lim \left(1+\frac{1}{\frac{n}{2}}\right)^{2\cdot\frac{n}{2}}=\lim \left(1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\cdot\lim \left(1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}$$

I'd say that those two factors converge against $e$, just not as fast. My explanation would be, that the sets $\mathbb N$ and $\{\frac{n}{2}:n\in \mathbb N\}$ have infinitely many elements in common.

So is it enough to just (state that and) add $$=\lim \left(1+\frac{1}{n}\right)^n\cdot \lim \left(1+\frac{1}{n}\right)^n=e\cdot e=e^2$$?

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  • $\begingroup$ Your argument is intuitively correct but not formally correct. In the formula $\lim (1 + (1/n))^{n} = e$ we assume that $n$ is a positive integer. In your calculation you replace $n$ by $n/2$ which may not be an integer all the time. So the logic breaks down. But the fact is that $\lim(1 + (1/n))^{n} = e$ as $n \to \infty$ even if $n$ is not an integer but a real number and hence everything works fine. $\endgroup$
    – Paramanand Singh
    May 22, 2014 at 13:45
  • $\begingroup$ It is better to start with the general sequence $a_{n} = (1 + (x/n))^{n}$ and show that $\lim a_{n}$ exists for all $x$ and thus this limit defines a function of $x$, say $f(x)$. Then by definition $f(1) = e$ and we establish $f(x + y) = f(x)f(y)$ so that $f(2) = f(1)f(1) = e^{2}$. This approach is presented at paramanands.blogspot.com/2014/05/… $\endgroup$
    – Paramanand Singh
    May 22, 2014 at 13:55

2 Answers 2

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Your substitution is valid: if $u(x)$ is monotic increasing then $\displaystyle\lim_{x\to\infty} f(x)=\lim_{x\to\infty} f(u(x))$. However as pointed out by Ian, this has nothing to do with a $\Bbb N$ and $u(\Bbb N)$ having an infinite number of elements in common. It is fairly straightforward to find an $f$ and $u$ where $\Bbb N\cap u(\Bbb N)$ is infinite but the limit $\lim f$ does not exist where $\lim f\circ u$ does. Try to find one! (However if the first exists, then the second must also exist and be equal to it. Can you prove this too?)

If $\lim a_n$ and $\lim b_n$ exist (and are finite), then $\lim (a_nb_n)$ also exists and equals $(\lim a_n)(\lim b_n)$.

Alternatively, since the map $x\mapsto x^2$ is continuous, we have $\lim a_n^2=(\lim a_n)^2$, more direct.

All of these facts fall under the heading of real analysis. Are you familiar with these facts? If so, then your argument is good. Otherwise, perhaps familiarize yourself with some basic analysis.

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    $\begingroup$ The substitution is indeed valid, but the arguments in OP are not. It has nothing to do with the fact that sets "have an infinite number of elements in common". $\endgroup$
    – Ian
    May 21, 2014 at 3:32
  • $\begingroup$ @Ian Ah, I had skimmed over that in the OP's question. $\endgroup$
    – anon
    May 21, 2014 at 3:36
  • $\begingroup$ So u(x)=x/2? @Ian I know that it would not work in R, but why not in N? $\endgroup$ May 21, 2014 at 4:02
  • $\begingroup$ @Carlster, take $f(n)=n\xrightarrow[n\rightarrow\infty]{}+\infty$, and $f((-1)^n\frac n2)$ does not have a limit whereas $\mathbb N$ and $\{(-1)^n\frac n2\ :\ n\in\mathbb N\}$ have an infinite number of elements in common. $\endgroup$
    – Ian
    May 21, 2014 at 4:34
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Just for your curiosity, the problem can also be approached using Taylor series; for large values of $n$, can be established $$\left(1+\frac{a}{n}\right)^n=e^a-\frac{a^2 e^a}{2 n}+\frac{a^3 (3 a+8) e^a}{24 n^2}-\frac{a^4 (a+2) (a+6) e^a}{48 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

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