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I know the proof using binomial expansion and then by monotone convergence theorem. But i want to collect some other proofs without using the binomial expansion.

*if you could provide the answer without the concept of calculus it will be appreciated * But any way do provide whatever method you can.

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    $\begingroup$ It is hard to avoid "the concept of calculus" since limits and convergent sequences are a part of that concept. On the other hand, it would help to specify what tools you're happy with using, since this result is used in developing some of them. (For example, if you define $e^x = \lim_{n \to \infty} (1 + x/n)^n$, then clearly we should not be using $e^x$ in the process of showing that this limit exists.) $\endgroup$ – Misha Lavrov Mar 27 '19 at 17:24
  • $\begingroup$ math.stackexchange.com/questions/389793/… $\endgroup$ – marty cohen Mar 27 '19 at 17:56
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If $a_n = \left(1 + \frac{1}{n}\right)^n$, then $\ln(a_n) = n \cdot \ln\left(1+\frac{1}{n}\right)$. This is indeterminate, but you can apply l'Hopital's rule to find $\ln(a_n)\to 1$ with $n$, so that, using the fact that continuous functions preserve limits, $a_n\to e$.

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  • $\begingroup$ Is their a way without using the concept of calculus? $\endgroup$ – Rkb Mar 27 '19 at 17:16
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Let $a_n$ the given sequence. We have

$$w_n:=\ln(a_{n+1})-\ln a_n=\ln\frac{a_{n+1}}{a_n}=(n+1)\ln(1+1/(n+1))-n\ln(1+1/n)=O(\frac1{n^2})$$ so the series $\sum w_n$ is convergent and so by telescoping the sequence $(\ln(a_n))$ is convergent. Conclude using the exponential function.

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You can do this in two steps; here, no calculus is needed:

Let $a_n=(1+\frac1n)^n$.

  1. Show that $a_n\leq\sum_{k=0}^n\frac1{k!}$ for all $n\in\mathbb N$.
  2. Show that $a_n$ is monotonically increasing.

To show (2), you have to prove that $\frac{a_{n+1}}{a_n}\geq1$; some calculations and Bernoulli's inequality are involved here:

$$\frac{a_{n+1}}{a_n}=\frac{\left(1+\frac{1}{1+n}\right)^{n+1}}{\left(1+\frac1n\right)^n}=\left(1+\frac1n\right)\left(\frac{1+\frac{1}{1+n}}{1+\frac1n}\right)^{n+1}=\left(1+\frac1n\right)\left(\frac{n^2+2n}{n^2+2n+1}\right)^{n+1}=\left(1+\frac1n\right)\left(1-\frac{1}{n^2+2n+1}\right)^{n+1}\geq\left(1+\frac1n\right)\left(1-\frac{n+1}{n^2+2n+1}\right)=\frac{n+1}{n}\frac{n}{n+1}=1$$

Because the series on the right hand side in (1) is convergent, $(a_n)$ is bounded. Together with (2), this implies that the sequence converges.

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  • $\begingroup$ 1 is easy, 2 is not. $\endgroup$ – marty cohen Mar 27 '19 at 17:55
  • $\begingroup$ Edited my answer to include the calculations for (2). $\endgroup$ – st.math Mar 27 '19 at 18:08
  • $\begingroup$ Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1. $\endgroup$ – marty cohen Mar 27 '19 at 21:09
  • $\begingroup$ See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/… $\endgroup$ – marty cohen Mar 27 '19 at 21:13
  • $\begingroup$ You are right, thanks. :) Corrected it. $\endgroup$ – st.math Mar 27 '19 at 22:02

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