Mathematica knows that:

$$n^k=\lim_{s\to 1} \, \frac{\zeta (s) \left(1-\frac{1}{\exp ^{s^{n^k}-1}(n)}\right)}{n}$$

Why is the above a trivial identity? What is it about the Zeta function that makes it obvious?

I know experimentally that one can test a zeta zero with the integral:

$$\int_0^{\infty } \frac{1}{\exp \left(x^{\frac{1}{\rho _1}}\right)+1} \, dx$$

which resembles the expression inside the parentheses in the limit a little bit.

If any one knows how to rewrite the latex of the limit to make it more readable, feel free to edit.

As a Mathematica program this limit is:

Clear[s, n]
Limit[Zeta[s]*(1 - 1/Exp[n]^(s^n^k - 1))/n, s -> 1]
  • $\begingroup$ Can the n^k be replaced by any function? $\endgroup$ – Mats Granvik Jul 16 '13 at 12:33
  • $\begingroup$ On which domain are you with $s$? $\endgroup$ – al-Hwarizmi Jul 16 '13 at 13:06
  • $\begingroup$ I think I am on the reals. Is that right? s goes towards 1. The number 1 is an integer. I don't know domains that well. $\endgroup$ – Mats Granvik Jul 16 '13 at 13:12
  • $\begingroup$ See reference.wolfram.com/mathematica/ref/Zeta.html $\endgroup$ – al-Hwarizmi Jul 16 '13 at 13:59

$$\tag11-\frac1{\exp(n)^{s^{n^k}-1}}=1-\exp(n(1-s^{n^k}))$$ where $1=\exp(n(1-1^{n^k}))$, so $$ \lim_{s\to 1}\frac1{s-1}\left(1-\frac1{\exp(n)^{s^{n^k}-1}}\right)$$ is by definition the derivative of $(1)$ at $s=1$, which is $$\left.\frac d{ds}\left(1-\exp(n(1-s^{n^k}))\right)\right|_{s=1}=\left.-n(-n^k)s^{n^k-1}\exp(n(1-s^{n^k}))\right|_{s=1} =n^{k+1}$$ Now all we need to know about $\zeta$ is that $\lim_{s\to1}(s-1)\zeta(s)=1$ as everything nicely cancels.

  • $\begingroup$ Yep, thanks. The same correction was of course needed for the derivative of that complicated expression. $\endgroup$ – Hagen von Eitzen Jul 16 '13 at 17:48

Change variables to $s=u+1$ and we are then asked to prove: $$n^k=\lim_{u\to 0^{+}}\frac{\zeta(u+1)}{n}\left(1-e^{n(1-(u+1)^{n^k}})\right)$$ Note that: $$1- \left( u+1 \right) ^{{n}^{k}}=-\sum _{q=1}^{\infty }{{n}^{k} \choose q}{u}^{q}=-n^ku+... $$ and so: $$\lim_{u\to 0^{+}}e^{n(1-(u+1)^{n^k}})=\lim_{u\to 0^{+}}e^{-n^{k+1}u}=\lim_{u\to 0^{+}}(1-n^{k+1}u)$$

$$\lim_{u\to 0^{+}}\frac{\zeta(u+1)}{n}\left(1-e^{n(1-(u+1)^{n^k}})\right)=\lim_{u\to 0^{+}}\zeta(u+1)n^ku$$ It remains to prove:$$\lim_{u\to 0^{+}}\zeta(u+1)u=1$$ To do so we borrow two known results from analysis; the functional identity for the Riemann Zeta function: $$\zeta \left( u+1 \right) =2\,{\pi }^{u}\cos \left( 1/2\,\pi \,u \right) \Gamma \left( -u \right) \zeta \left( -u \right) {2}^{u} \tag{1}$$ and: $$\lim_{u=0^{+}}\zeta(u)=\lim_{u=0^{-}}\zeta(u)=\zeta(0)=-1/2\tag{2}$$ where $(2)$ follows by the continuity of $\zeta(u)$ away from the only pole at $\zeta(1)$ and the limiting value is established here. Multiplying $(1)$ by $u$ we have:

$$u\zeta \left( u+1 \right) =u2\,{\pi }^{u}\cos \left( 1/2\,\pi \,u \right) \Gamma \left( -u \right) \zeta \left( -u \right) {2}^{u}$$ which by $u\Gamma(-u)=-\Gamma(1-u)$ becomes: $$u\zeta \left( u+1 \right) =-2\,{\pi }^{u}\cos \left( 1/2\,\pi \,u \right) \Gamma \left( -u+1 \right) \zeta \left( -u \right) {2}^{u}$$ Then by the continuity of $\Gamma(u)$ over the positive reals and the fact that $\Gamma(1)=0!=1$ we have: $$\lim_{u=0^{+}}-2\,{\pi }^{u}\cos \left( 1/2\,\pi \,u \right) \Gamma \left( -u+1 \right) {2}^{u}=-2$$ which together with $(2)$ proves the limit: $$\lim_{u=0^{+}}u\zeta \left( u+1 \right)=-2\lim_{u=0^{+}}\zeta(-u)=1$$


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