If you consider the fact that: $$\lim_{n\to\infty} \left(1+{1\over n}\right)^n=e$$ Where $e$ is Euler's number. Now, the prime number theorem states that: $$\lim_{x\to\infty} {\pi(x)\over x / \ln{x}}=1$$ Where $\pi$ is the prime-counting function. Now, think about, if we say that: $$\pi(x)={x\over \log_{c_x}{x}}$$ Where $c_x$ is some constant at the $x$th order, and the prime-number theorem is equivalent to saying: $$\lim_{x\to\infty} c_x=e$$ Now, consider that to calculate $c_x$ based on $\pi(x)$ is quite easy: $$c_x=\sqrt[{x\over \pi(x)}]{x}=x^{\pi(x)\over x}=(\sqrt[x]{x})^{\pi(x)}$$ Now, the prime-number theorem can be written as: $$\lim_{x\to\infty} (\sqrt[x]{x})^{\pi(x)}=\lim_{n\to\infty}\left(1+{1\over n}\right)^n$$ We can see that $\lim_{x\to \infty} \sqrt[x]{x}=\lim_{n\to\infty}1+{1\over n}$, since both $\rightarrow 1$, and considering that both the powers diverge, this may be a stupid question, but, is this enough to be a proof of the prime-number theorem? If not, what more do you need here, for this to be a proof of it?
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@InterstellarProbe Here's a short proof that $lim_{x\to\infty} (1 + {1\over x})^x=e$: https://math.stackexchange.com/questions/882741/limit-of-1-x-nn-when-n-tends-to-infinity – Tots Jan 24 '20 at 15:29
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2Your proof contains the asumtion that $\pi(x)$ has this asymptotic behaviour. Just that this does not lead to a contradiction, does not prove that $\pi(x)$ actually behaves asymptotically this way. – Peter Jan 24 '20 at 15:36
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@Peter My proof only assumes that $\pi(x)\rightarrow \infty$ as $x\rightarrow\infty$. – Tots Jan 24 '20 at 15:42
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@InterstellarProbe I have edited my question. – Tots Jan 24 '20 at 16:40
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2Where have you used that $\pi(x)$ counts the prime numbers $\leq x>$? – Christian Blatter Jan 24 '20 at 16:53
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1@Tots As $n \to \infty$, $2n \to \infty$, but $\left(1+\dfrac{1}{n}\right)^{2n} \to e^2 \neq e$ as $n \to \infty$. Similarly, $n^n \to \infty$ as $n \to \infty$, but $$\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^{n^n} = \infty \neq e$$ So, as Peter says, not showing a contradiction is not the same as a proof. You have not shown anything about the asymptotic nature of $\pi(x)$. – SlipEternal Jan 24 '20 at 17:07
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Here's another issue: $$\lim_{x \to \infty} \dfrac{\sqrt[x]{x}}{1+\dfrac{1}{x}} = 1$$ but $$e = \lim_{x \to \infty}\left(1+\dfrac{1}{x}\right)^x \neq \lim_{x \to \infty} \left(\sqrt[x]{x}\right)^x = \lim_{x \to \infty} x = \infty$$ – SlipEternal Jan 24 '20 at 17:10
1 Answers
In your last paragraph you seem to be using the following assertion: if $\lim_{x\to\infty}f(x)=1$ and $\lim_{x\to\infty} g(x)= \infty$, then $\lim_{x\to\infty} f(x)^{g(x)} = e$. This is not true. Consider for example \begin{align*} \lim_{x\to\infty} \big( 1+\tfrac1x \big)^{\sqrt x} &= 1 \\ \lim_{x\to\infty} \big( 1+\tfrac1x \big)^{x/2} &= \sqrt e \\ \lim_{x\to\infty} \big( 1+\tfrac1x \big)^{x} &= e \\ \lim_{x\to\infty} \big( 1+\tfrac1x \big)^{3x} &= e^3 \\ \lim_{x\to\infty} \big( 1+\tfrac1x \big)^{x^2} &= \infty \end{align*} all of which can be proved by taking logarithms and using $0\times\infty$ indeterminate form techniques. (Indeed, such an $f(x)^{g(x)}$ is a $1^\infty$ indeterminate form; if the answer always equaled $e$, then it wouldn't be called an indeterminate form at all!)
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If we compare $(1+{1\over x})^x$ with $f(x)^{g(x)}$, I was thinking about if you proved, that $\lim_{x\to\infty} f(x)=1$, but $f(x)$ tends to $1$ more slowly than $1 + {1\over x}$, then $g(x)$'s growth rate difference with $x$ has to be similar to the difference of the growth rate between $f(x)$ and $1 + {1\over x}$, so that maybe, $lim_{x\to\infty} (1 + {1\over x})^x=lim_{n\to\infty}f(n)^{g(n)}$ – Tots Jan 24 '20 at 18:13
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1The correct way to resolve this "maybe" is to carry out the indeterminate-form process to find out what the answer is. In its inverse form as you proposed to use it, it can't tell the difference between $\pi(x) \sim x/\log x$ and $\pi(x) \sim x/(\log x)^{100}$ and even $\pi(x) \sim x(\log x)^{100}$. – Greg Martin Jan 25 '20 at 03:52