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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$\begingroup$You can do this in two steps; here, no calculus is needed:
Let $a_n=(1+\frac1n)^n$.
- Show that $a_n\leq\sum_{k=0}^n\frac1{k!}$ for all $n\in\mathbb N$.
- 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.
$\endgroup$ 5 $\begingroup$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, $a_n\to e$.
$\endgroup$ 0 $\begingroup$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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