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I'm very confused about the question below, which I couldn't figure out for days.

In Example 5 the author is teaching us proving

$$\lim_{n\to \infty} (1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\frac{1}{n!}) = e$$

with equation

$$x_n = (1+\frac{1}{n})^n$$ $$y_n = 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\frac{1}{n!}$$

I'm thinking $x_n < e$ if we take $e$ is the special value only $n$ tends to infinity, and term by term comparing tells us $x_n \le y_n$ if we consider $n$ can be $1$, then we have the following: $x_n \le y_n, x_n < e$, but I believe that doesn't give us $e\le y_n$. I think only if we use $x_n = e$ and with $x_n \le y_n$ brings us $e \le y_n$. So I'm quite confused how did him resulted $e \le e$ first?

Then $m < n$ is implemented to help us go further, with $x_m < x_n < e$ and resulting $y_m\le e$? How? I'm thinking such outcomes only come from $x_m \le y_m$ and $x_m < x_n \le e$.

I really need your help.

Page url: Calculus With Analytic Geometry By George Simmons

John HHU
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    Does this answer your question? – D S Mar 02 '24 at 20:07
  • First you must tell us your definition of $e$. For instance, some people define $e$ as the limit as $n$ tends to infinity of $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\frac{1}{n!}$. But then your question would answer itself! – TonyK Mar 02 '24 at 20:36
  • The definition of e is this: $\lim_{n\to \infty} (1+\frac{1}{n})^n = e$ – John HHU Mar 03 '24 at 00:59
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    See this application of Tannery’s theorem. – KStarGamer Mar 03 '24 at 03:57
  • Thank you but are they any intuitive approach understanding such a proving process? The Tannery’s theorem is harder for me without understanding many base knowledge. – John HHU Mar 03 '24 at 04:36
  • Expanding $x_{n}$ using binomial theorem for any index and taking the limit of $n$ as it tends to infinity gives $y_{n}$ I think. – AAM Mar 03 '24 at 05:35
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    My lecture notes on the exponential function have a proof for this. I did it for the sake of completeness, because the same concern occurred to me, also. The catch is that those lecture notes are in Finnish. The proof you want is that of Lause 0.6., straddling pages 4 and 5. The ingredients I use: A) the binomial theorem, B) definition of $\exp(x)=\sum_{n}x^n/n!$, C) the functional equation $\exp(x+y)=\exp(x)\exp(y)$, D) the definition $$e=\lim_{n\to\infty}(1+\frac1n)^n.$$ – Jyrki Lahtonen Mar 03 '24 at 06:07
  • Thank you for your comments, to @AAM, the expanding of $x_n$ will give you extra terms $(1-\frac{1}{n})*(1-\frac{2}{n})...$, which will make $x_n <= y_n$. – John HHU Mar 03 '24 at 06:45
  • Thank you very much @JyrkiLahtonen, the equation in page 5, how could we for sure that

    "when $n > k > 0$ and $n$ tends to infinity will swipe the $-i$ influence which goes from 0 to k"?

    Cause I'm thinking as each term multiplied a term which is less than 1 before read your note

     – John HHU Mar 03 '24 at 07:27
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    @John On top of page 5 I fix $k$ (for a few lines), and then let $n\to\infty$. Then for all $i$ in the range $0<i\le k$ we have $(n-i)/n\to1$, and arrive at a lower bound $$e\ge\sum_{i=0}^k\frac1{i!}$$ that holds first for this fixed value of $k$, but because the choice was arbitrary it holds for all $k$. – Jyrki Lahtonen Mar 03 '24 at 09:38
  • @JyrkiLahtonen Sorry I put many confusing comments here before. How about page 4 last equation, the value $k$ made $(1+\frac{1}{n})^n\ge\lim_{n\to \infty}\sum_{k=0}^m\frac{1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})}{k!}$? Why its $\ge$ sign instead of just a $>$ sign? – John HHU Mar 04 '24 at 16:38

1 Answers1

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Let $x_n=(1+\frac{1}{n})^n.$By the binomial theorem, we have

$x_{n}=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})}{k!}$

Find an integer $m$ such that $m<n$,there is

$x_n>\sum_{k=0}^m\frac{1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})}{k!}$

Let $n\rightarrow \infty$,then $\mathrm{e}\geq \sum_{k=0}^m\frac{1}{k!}$.Similarly, let $m\rightarrow \infty$,we can get $\mathrm{e}\geq \sum_{n=0}^\infty\frac{1}{n!}$

On the other hand,

$x_{n}=\sum_{k=0}^n\frac{1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})}{k!}\leq\sum_{k=0}^n\frac{1}{k!}$

Let $n\rightarrow \infty$ ,we get $\mathrm{e}\leq \sum_{n=0}^\infty\frac{1}{n!}$.Hence , by the sandwich theorem,we have

$\mathrm{e}=\sum_{n=0}^\infty\frac{1}{n!}$

RainField
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  • Thank you very much @RainField, I was following the pdf note above and couldn't let it go. I saw the similarity of both answers, and I guess its standard procedure in Mathematical analysis, but why can we step further from $m < n$ with $n\to\infty$ to $m\to\infty$? – John HHU Mar 03 '24 at 14:19
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    For the expansion of $x_n$, we can see the limit of numerator is $1$ when $n\rightarrow\infty$. Then the rest of fraction is $\frac{1}{k!}$,which is what we want.But as you can see, when we let $n\rightarrow\infty$, the number of items also increases to infty, which isn't what we want. So we hope that we can control the number of items, that's the reason why $m$ appears. – RainField Mar 03 '24 at 15:04
  • May I ask why its $\geq$ sign instead of $>$ sign in $\mathrm{e}\geq \sum_{k=0}^m\frac{1}{k!}$? – John HHU Mar 05 '24 at 04:48
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    Because it's not sure whether you can get $\lim a_n> \lim b_n$ as $n\rightarrow\infty$ from $a_n>b_n$.For example,$\frac{1}{n}$ and $\frac{1}{n+1}$. – RainField Mar 09 '24 at 09:02
  • Thank you, so $\mathrm{e}\geq \sum_{k=0}^m\frac{1}{k!}$ is based on 1) the possibility that $m = n-1$ and 2) the $n\to \infty$ gives us $1\cdot(1-\frac{1}{n})\cdots(1-\frac{k-1}{n}) \to 1$, and that's where $m$ comes to help. Thank you very much – John HHU Mar 12 '24 at 04:21