I have already proven that $\left(1+\frac 1{n}\right)^n\leq \sum_{k=0}^{n}{\frac{1}{k!}}$ . Does that help me in any way? I am stumped...
Any suggestions/hints?
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Note that your last equality is not true: $\sum\limits_{k=0}^n\frac{1}{k!}\neq e.$ – cmk Jun 07 '19 at 18:32
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updated. regards, para. – Analysis Jun 07 '19 at 18:33
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https://math.stackexchange.com/questions/254335/prove-that-5-2-e-3/254339#254339 – lab bhattacharjee Jun 07 '19 at 18:35
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You can expand the left hand side using the binomial theorem. – mjw Jun 07 '19 at 18:35
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Are we talking about $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ – Analysis Jun 07 '19 at 18:37
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You can calculate the limit of the log. – Bernard Jun 07 '19 at 18:38
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This is often taken as the definition of $e$. So what are your premises ? – Jun 07 '19 at 18:42
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How do I prove $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^{\infty}{\frac{1}{k!}}$, that's my question. – Analysis Jun 07 '19 at 18:52
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Do you just need to prove the limit equals that sum, or also that the sum equals $e$. – eyeballfrog Jun 07 '19 at 19:02
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Solely that the limit equals the sum. – Analysis Jun 07 '19 at 19:02
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basically, why this is true: https://www.wolframalpha.com/input/?i=sum_(k%3D0)%5E%E2%88%9E+1%2F(k!)+%3D+lim_(n-%3E%E2%88%9E)+(1+%2B+1%2Fn)%5En – Analysis Jun 07 '19 at 19:04
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An "interesting" take at this. – Jyrki Lahtonen Jun 07 '19 at 19:33
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Done here too https://math.stackexchange.com/a/2172435/399263 – zwim Jun 07 '19 at 21:37
2 Answers
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If $n,N\in\mathbb N$, with $n\geqslant N$, then\begin{align}\left(1+\frac1n\right)^n&=\sum_{k=0}^n\binom nk\frac1{n^k}\\&\geqslant\sum_{k=0}^N\binom nk\frac1{n^k}\\&=1+1+\frac{n-1}{2n}+\frac{(n-1)(n-2)}{3!n^2}+\cdots+\frac{(n-1)(n-2)\ldots\bigl(n-(N-1)\bigr)}{N!n^{N-1}}\\&=1+1+\frac1{2!}\left(1-\frac1n\right)+\frac1{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots\\&\phantom{=\ }+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{N-1}n\right).\end{align}Therefore,\begin{align}\left(1+\frac1n\right)^n&\geqslant\lim_{n\to\infty}1+1+\cdots+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\frac1{2!}\left(1-\frac1n\right)\cdots\left(1-\frac{N-1}n\right)\\&=1+1+\frac1{2!}+\cdots+\frac1{N!}\end{align}and so$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\geqslant1+1+\frac1{2!}+\cdots+\frac1{N!}.$$Since this takes place for every $N\in\mathbb N$,$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\geqslant\sum_{k=0}^\infty\frac1{k!}.$$
José Carlos Santos
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$\begin{align}\left(1+\frac1n\right)^n&\geqslant\lim_{n\to\infty}1+1+\cdots+\frac1{N!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\frac1{2!}\left(1-\frac1n\right)\cdots\left(1-\frac{N-1}n\right)\&=1+1+\frac1{2!}+\cdots+\frac1{N!}\end{align}$ How do you get that Inequality? – Analysis Jun 08 '19 at 11:04
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How do I know what? Are you talking about the first inequality of my proof? – José Carlos Santos Jun 08 '19 at 11:36
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I don't know if this is what you're looking for but:
let exp(x) be the function that satisfies the D.E.: $$\frac{d}{dx}\exp(x) = \exp(x),\qquad \exp(0) = 1$$ hence if we do a Taylor expansion of this function we get: $$\exp(x) = \sum_{k=0}^\infty{\frac{x}{k!}}$$ using the limit definition of the derivative with a constant to the power x: $$\frac{d}{dx}\left(k^x\right) = \lim \limits_{h \to 0}\left(\frac{k^{x+h} - k^{x}}{h}\right) = \lim \limits_{h \to 0}\left(k^{x}\left(\frac{k^{h} - 1}{h}\right)\right) = T\cdot k^{x} $$ where $T = \lim \limits_{h \to 0}\left(\frac{k^{h} - 1}{h}\right)$ is constant with respect to x. Let e be the number such that $T = 1$, trivially, $e^x = \exp(x)$. rearranging our formula for T: $$\lim \limits_{h \to 0}\left[\frac{e^{h} - 1}{h}\right] = 1 \Rightarrow \lim \limits_{h \to 0}\left[e =\left(1+h\right)^\frac{1}{h}\right]$$ letting $u = \frac{1}{h}$: $$e =\lim \limits_{u \to \infty}\left[\left(1+\frac{1}{u}\right)^u\right]$$ hence: $$\lim \limits_{x \to \infty}\left[\left(1+\frac{1}{x}\right)^x\right] = e = \exp(1) = \sum_{k=0}^\infty{\frac{1}{k!}}$$
S. Dauncey
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