Is there any online source for proofs than $(1+\frac{1}{n})^n$ and $(1+\frac{1}{x})^x$ is $e$?

Question

Can someone recommend a good online source where it is proved that the limit for sequence $(1+\frac{1}{n})^n$ as $n$ tends to infinity and function $(1+\frac{1}{x})^x$ as $x\rightarrow \infty$ is e? Most appreciated would be detailed proofs so that newbies can understand it as well. Where the boundedness and monotony would be proved for the sequence. Also proofs that use l'hospital rule don't work for me, as I need the proof for the exam where the subject of derivatives is not yet been taught. I have searched on the internet, but haven't come across a good proof, so if you know where to find a really good proof with explanation, please give me a note!

Answer 1

I think this is the proof Dr Sonnhard Graubner is talking about: Prove $(1 + \frac{1}{n})^n$ is bounded above (don't have 50 rep, so can't post it as comment)

Answer 2

Define the sequence $$a_n=\left(1+\frac1n\right)^n$$ Note that, using the binomial theorem, we have that $$a_n=\sum_{r=0}^n\binom{n}{r}\frac1{n^r}$$ $$a_{n+1}=\sum_{r=0}^{n+1}\binom{n+1}{r}\frac1{(n+1)^r}$$ Hence $$a_{n+1}-a_n=\sum_{r=0}^n\left[\binom{n+1}{r}\frac1{(n+1)^r}-\binom{n}{r}\frac1{n^r}\right]+\frac1{(n+1)^{n+1}}$$ Now, $$\begin{align} \binom{n}{r}\frac1{n^r} &=\frac{n(n-1)\cdots(n-r+1)}{r!n^r}\\ &=\frac1{r!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{r-1}n\right)\\ \end{align}$$ and similarly $$\binom{n+1}{r}\frac1{(n+1)^r}=\frac1{r!}\left(1-\frac1{n+1}\right)\left(1-\frac2{n+1}\right)\cdots\left(1-\frac{r-1}{n+1}\right)$$ Each factor in the latter product is greater than the corresponding factor in the earlier product. Hence we can deduce that $a_{n+1}-a_n\gt0$ and so $a_n$ is strictly monotonic increasing. One can also show that this sequence is bounded above and hence converges by the Monotone Convergence theorem. The value to which this sequence converges is denoted by $e$.