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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!

Martin Sleziak
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user
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    Isn't that the definition of $e$ though? Definitions can't be "proven" unless you have defined it some other way –  Aug 27 '19 at 15:17
  • You mean, proof that the limit exists? – Vasili Aug 27 '19 at 15:17
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    See here https://math.stackexchange.com/questions/2642478/proof-of-e-as-a-limit – Dr. Sonnhard Graubner Aug 27 '19 at 15:18
  • I think OP would be happy with a proof that both limits equal $e = \sum_{n=0}^\infty\tfrac 1{n!}$. – amsmath Aug 27 '19 at 15:22
  • @Dr.SonnhardGraubner The proof there includes derivatives, so it doesn't suite me. – user Aug 27 '19 at 16:02
  • @Vasya Ok, I made an error assuming it is possible to prove that the limit of this sequence is as there is no proof for that. But there is a proof of sequence being monotonic and bounded. That's what I'd like to find, and a proof with l'hospital rule or derivatives don't fit me. As for function I think there must be a proof where the function is bounded by this sequence so it must have the same limit or smth similar. – user Aug 27 '19 at 16:06
  • @user: I've seen a proof which is based on binomial expansion. The proof shows that the limit of that sequence is between $2$ and $3$. I may post it later if I have time. – Vasili Aug 27 '19 at 16:40
  • @Vasya Ok, thank you that would be awesome. ;) – user Aug 27 '19 at 16:46

2 Answers2

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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)

DeDeen
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  • @drsonhardgraubner Such proof doesn't suit me, because it uses derivatives, I need a proof where the existence of the limit is derived from the boundedness of the sequence and it's monotonic nature. – user Aug 27 '19 at 18:01
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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 $e$.

Peter Foreman
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  • Thank you for proving that it is increasing sequence, but where could I possibly find a proof of it being bounded by 2 an 3? Also I need a proof of the limit of the function when x tends to infinity, maybe you have a great source? – user Aug 27 '19 at 20:07
  • Note that $a_n\lt\sum_{r=0}^n\frac1{r!}\le2+\sum_{r=2}^n\frac1{2^{n-1}}$ for $n\ge1$ which can be used to deduce the upper bound of $3$. The lower bound of $2$ is clear from $a_1=2$ and the increasing nature. – Peter Foreman Aug 27 '19 at 20:11
  • Why is that $\sum_{r=0}^n\frac{1}{n^r}<\sum_{r=0}^n\frac{1}{r!}$ and I can't make the sense of the last sum, why it is important to separate these first two terms from it? – user Aug 27 '19 at 20:32
  • It's not $\sum_{r=0}^n\frac1{n^r}$ it's $a_n=\sum_{r=0}^n\binom{n}{r}\frac1{n^r}$ note that the terms of the sum for $a_n$ are equal to $\frac1{r!}(1-1/n)(1-2/n)\cdots(1-(r-1)/n)\lt\frac1{r!}$ hence $a_n\lt\sum_{r=0}^n\frac1{r!}$. For the second bound we require $2^{r-1}\gt r!$ (I meant to write $\sum_{r=2}^n\frac1{2^{r-1}}$) and hence $r\ge2$, the first two terms are seperately evaluated and hence the inequality holds. – Peter Foreman Aug 27 '19 at 20:41
  • Ok, but the solution should be that $2<a_n<3$, how to show that? – user Aug 27 '19 at 21:10
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    I literally stated how to find both bounds in my initial comment. We have $a_1=2$ hence $a_n\gt2$ then the upper bound of $3$ follows from $a_n\lt2+\sum_{r=2}^\infty\frac1{2^{r-1}}=3$. – Peter Foreman Aug 27 '19 at 21:20