Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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]
share|cite|improve this question
Can the n^k be replaced by any function? – Mats Granvik Jul 16 '13 at 12:33
On which domain are you with $s$? – al-Hwarizmi Jul 16 '13 at 13:06
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. – Mats Granvik Jul 16 '13 at 13:12
up vote 3 down vote accepted

$$\tag11-\frac1{\exp(n)^{s^{n^k}-1}}=1-\exp(n(1-s^{n^k}))$$ 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.

share|cite|improve this answer
Yep, thanks. The same correction was of course needed for the derivative of that complicated expression. – 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: $$\lim_{u=0^{+}}u\zeta \left( u+1 \right)=-2\lim_{u=0^{+}}\zeta(-u)=1$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.