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If I already know that $$\lim \limits_{n \to \infty} a_n=+\infty$$ Then how can I prove $$\lim \limits_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e$$ without involving function limit?

This question comes because you may find some books on calculus or analysis (maybe they are badly written) require you to prove something like $$\lim \limits_{n \to \infty}\left(1+\frac{2}{n}\right)^{n}=e^2$$ or something more complex even before they formally introduce the definition of limit of a function, They are hard to prove because you can't simply take something like $\frac{n}{2}$ as a subsequence of $n$.

The definition of limit of a function (at infinity) here mean:

For a real function $f$ which is well-defined on $[a, +\infty)$, if for any $\epsilon >0$, there is a positive number $M \geq a$ such that when $x>M$ we can say $|f(x)-A|<\epsilon$, then $$\lim \limits_{x \to \infty}f(x)=A.$$

While the definition of limit of a sequence here mean:

For a sequence $\{a_n\}$, if for any $\epsilon>0$, there is a positive integer $N$ such that when $n>N$ we can say $|a_n-A|<\epsilon$, then $$\lim \limits_{n \to \infty}a_n=A.$$

I know sequence is a "special" kind of function whose domain is $\mathbb{N}$ and thus sequence limit is but a special case of function limit. Here I say avoid involving the idea of function limit means not to use the idea above but only to prove it by the "special case" below. After all, $(1+\frac{1}{a_n})^{a_n}$ is still a "special" function - a sequence.

p.s. $e$ is defined by $$\lim \limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e.$$

My try so far:

Since $$\lim \limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e,$$ for every $\epsilon>0$, there is $N \in \mathbb{N}$ s.t. for all $n>N$,$$|\left(1+\frac{1}{n}\right)^n-e|<\epsilon.$$

Meanwhile, since $$\lim \limits_{n \to \infty} a_n=+\infty,$$ for $N' \in \mathbb{N}$ and $N'>N$, there is $N'' \in \mathbb{N}$ s.t. for all $n>N''$, $a_n>N'>N.$

However, if $a_n$ become bigger then $1+\frac{1}{a_n}$ will be smaller, and vise versa, so I don't know how to deal with $\left(1+\frac{1}{a_n}\right)^{a_n}.$

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  • $\begingroup$ What's your definition of $e$? Often it is $\lim_{n\to\infty}\left(1+\frac1n\right)^n$ ... $\endgroup$ – Hagen von Eitzen Sep 25 '15 at 12:59
  • $\begingroup$ @HagenvonEitzen Oh I forgot to point it out. It is the definition of $e$ I'm dealing with. $\endgroup$ – Asydot Sep 25 '15 at 13:01
  • $\begingroup$ Sum of inverse factorials? en.wikipedia.org/wiki/E_(mathematical_constant) $\endgroup$ – UXkQEZ7 Sep 25 '15 at 13:05
  • $\begingroup$ It would help if you could explain what you mean by "function limit" and "function idea." Those are ambiguous terms, so it is hard to proceed without using them if we have no idea what you mean. $\endgroup$ – Thomas Andrews Sep 25 '15 at 13:13
  • $\begingroup$ @ThomasAndrews I think I fixed it. I just mean the definition and properties of the limit of a function. With the definition of limit of a function I think it can be proved easily. $\endgroup$ – Asydot Sep 25 '15 at 13:18
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Any subsequence of a Cauchy sequence is a Cauchy sequence with the same limit point, hence $$ \lim_{n\to +\infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e $$ as soon as $\{a_n\}_{n\in\mathbb{N}}$ is a diverging sequence of natural numbers. The very last assumption can be dropped by noticing that $\frac{\log(1+x)}{x}$ is a positive, decreasing and convex function on $[0,1]$, hence if $a_m\in\mathbb{R}$ is between $n$ and $n+1$, $$\left(1+\frac{1}{a_m}\right)^{a_m}$$ is between $\left(1+\frac{1}{n}\right)^n$ and $\left(1+\frac{1}{n+1}\right)^{n+1}$, so the same conclusion as above follows by squeezing.

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  • $\begingroup$ Sorry for I didn't get it... for example I given, $\frac{n}{2}$ can't be a subsequence of $n$. $\endgroup$ – Asydot Sep 25 '15 at 13:28
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    $\begingroup$ @Jack D'Aurizio:+1,Brilliant idea,Bravo! $\endgroup$ – Arpit Kansal Sep 25 '15 at 13:31
  • $\begingroup$ @Asydot The argument goes on to show that $a_{n}$ doesn't have to be a subsequence of $n$, because you can squeeze it close to one. $\endgroup$ – preferred_anon Sep 25 '15 at 13:32
  • $\begingroup$ @DanielLittlewood Oh, I see it. That is brilliant - however, is I am able to prove $\frac {\log (1+x)}{x}$ decreasing without derivative? $\endgroup$ – Asydot Sep 25 '15 at 14:06
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    $\begingroup$ @Asydot: you may try to prove it through the AM-GM or Bernoulli inequality. If you perform a search I am sure you will find many illuminating answers here on MSE about that point. $\endgroup$ – Jack D'Aurizio Sep 25 '15 at 16:01
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For $$\displaystyle\lim_{n\to\infty }\left(1+\frac{k}{n}\right)^n=e^k$$ with $k\in\mathbb N$, do it by induction. The case $k=1$ is the definition. For $k>1$, you have that $$\left(1+\frac{k+1}{n}\right)^n=\left(\frac{n+k+1}{n}\right)^n=\left(1+\frac{k}{n}\right)^n\left(1+\frac{1}{n+k}\right)^n=\underbrace{\left(1+\frac{k}{n}\right)^n}_{\to e^k\ (hyp\ induction)}\underbrace{\left(1+\frac{1}{n+k}\right)^{n+k}}_{\to e}\underbrace{\left(1+\frac{1}{n+k}\right)^{-k}}_{\to 1}\underset{n\to\infty }{\longrightarrow }e^{k+1}$$

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You can also evaluate the limit with l'Hospitals rule. The Limit $\big(1+\frac{1}{a_n}\big)^{a_n}$ with $\lim_{n\to\infty} a_n=+\infty$ is a limit of the type $1^\infty$. As a replacement for $a_n\to\infty$ I would like to use $k\to\infty$. With this you get: $$\lim_{k\to\infty} \big(1+\frac{1}{k}\big)^{k} = \lim_{k\to\infty} \mathrm{e}^{\ln\bigg(\big(1+\frac{1}{k}\big)^{k}\bigg)} = \lim_{k\to\infty} \mathrm{e}^{k\ln\big(1+\frac{1}{k}\big)}$$ Since the exponential function is a continous function, you can apply a limit theorem on the last expression. $$\lim_{k\to\infty} \big(1+\frac{1}{k}\big)^{k} = \mathrm{e}^{\lim_{k\to\infty} k \ln\big(1+\frac{1}{k}\big)}$$ Inside of the exponential function you get a limit of the type $\infty\cdot 0$ which can be transformed to a limit of the type $\frac{0}{0}$ by doing the following: $$\lim_{k\to\infty} \big(1+\frac{1}{k}\big)^{k} = \mathrm{e}^{\lim_{k\to\infty} \frac{\ln\big(1+\frac{1}{k}\big)}{\frac{1}{k}}}$$ You can apply l'Hospitals rule on the inner limit and you will get $1$ as result. So the limit will be $\mathrm{e}^1$.

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