9

By considering the integral Zeta function

$$F(s)=s+\frac{1}{2^s\ln(2)}+\frac{1}{3^s\ln(3)}+\frac{1}{4^s\ln(4)}+...$$

Evaluate

$$\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$$

EDIT:

There has clearly been much confusion here. I am asking for the analytic continuation of the integral Zeta function at 0. I am asking for the sum of the series in the sense that

$$1+2+3+...=-\frac{1}{12}$$

Gottfried Helms
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Elie Bergman
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  • No.${}{}{}{}{}$ –  Mar 29 '14 at 22:46
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    This is not homework in case your interested. I know the solution to this will probably involve some complex analysis which is beyond my knowledge at the moment. Hence my question. – Elie Bergman Mar 29 '14 at 23:07
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    F(s)=2^(-s)/ln(2)+3^(-s)/ln(3)+4^(-s)/ln(4)+... at s=0, The integral of the Riemann Zeta function. – Elie Bergman Mar 29 '14 at 23:11
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    Maybe you should put "regularization" in the title. Everybody is just repeating how obvious the divergence is. – orion Mar 29 '14 at 23:23

4 Answers4

11

This sum is of course divergent (as proved by others).

But the idea is probably to use zeta regularization and get a formal derivation like : \begin{align} \int_0^{\infty}(\zeta(x)-1)\,dx&=\int_0^{\infty} \sum_{n=2}^\infty \frac 1{n^x}\,dx\\ &=\sum_{n=2}^\infty \int_0^{\infty} e^{-x\ln(n)}\,dx\\ &=-\sum_{n=2}^\infty \left.\frac1{\ln(n)\,n^x}\right|_{x=0}^{\infty}\\ &=\sum_{n=2}^\infty \frac1{\ln(n)}\\ \end{align} From the simple pole of $\zeta(x)$ at $x=1$ we should write our 'regularized sum' as the Cauchy principal value of the integral : $$\sum_{n=2}^\infty \frac1{\ln(n)}=PV \int_0^{\infty}(\zeta(x)-1)\,dx\\=-0.243238342890980755415059\cdots$$ (if I didn't make an error...)

For numerical evaluation use $\displaystyle \lim_{N\to+\infty}\int_0^N \zeta(x)-1-\frac 1{x-1}\,dx+\ln(N-1)$.

Raymond Manzoni
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  • Hmm, the term $\zeta(x)- {1 \over x-1}$ occured so often lately in my exercises that I think an "incomplete zeta" function should become some standard function in the handbooks (and software). For such an "incomplete zeta" there is also a power series nicely converging at small absolute values of the argument... – Gottfried Helms Mar 30 '14 at 08:35
  • @GottfriedHelms: ... so that 'Stieltjes function' could be appropriate! :-) (Stieltjes studied in great detail $;\displaystyle\zeta(x)-\frac 1{x-1}$ and $\displaystyle\int \frac {\zeta(x)-\frac 12-\frac 1{x-1}}x,dx;$ after all). This function is analytic everywhere with just the (important) drawback that the zeros don't appear so neat. Cheers, – Raymond Manzoni Mar 30 '14 at 10:04
  • Nice answer. Do you suspect a closed form exists? In the case of the differential Zeta function, evaluating ln(2)+ln(3)+ln(4)+...=ln(sqrt(2*pi)). So one would suspect a closed form for the reciprocals also exists. – Elie Bergman Mar 30 '14 at 10:25
  • It's quite funny how about 4-5 people seem to answer all of my MSE questions. Small world! – Elie Bergman Mar 30 '14 at 10:41
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    @ElieBergman: I didn't find a closed form yet (using the $300$ first digits) and wouldn't bet on its existence (concerning the series many will return a closed form like $\sum_{n=2}^\infty \zeta(n)-1=1$ but integrals look harder perhaps because $\zeta(s)$ is often multiplied by $\Gamma(s)$ not sure...). Concerning the small world you are meeting people attracted by $\zeta$ and divergent series in these cases. Excellent continuation! – Raymond Manzoni Mar 30 '14 at 11:24
3

$\ln(x)$ is smaller than $x$ for all positive values.

By that logic we can say that $\frac{1}{\ln(x)}$ is larger than $\frac{1}{x}$ for all positive x's

This means that $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is larger than $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$

We know that $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$ diverges and equals an infinite sum, and since $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is larger than that infinite sum, we can conclude that $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$ is an infinite sum.

mle
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Asimov
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1

$$\sum^\infty_{n=1}\dfrac{1}{\ln(n+1)}$$ The comparison test, available here, shows that this is divergent.

0

Since $$ \ln k \le k \quad\forall k \ge 2, $$ it follows that $$ \sum_{k=2}^\infty\frac{1}{\ln k}\ge \sum_{k=2}^\infty\frac{1}{k}=\infty. $$

HorizonsMaths
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  • I don't really understand why my answer was downvoted. Could you please tell me where my mistake is! – HorizonsMaths Mar 29 '14 at 22:55
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    I think I have confused everyone in this question. I am asking for the analytic continuation of the sum. Your argument is correct, but I am asking for something completely different. – Elie Bergman Mar 29 '14 at 23:00
  • I edited it because as it stands your proof isn't correct, $-2^k \leq k$ $\forall k \geq 2$ but obviously $-\sum_{k = 2}^{\infty} \frac {1} {2^k}$ does exist. You need that ln$ k$ is positive as well for it to actually follow. – CameronJWhitehead Mar 30 '14 at 14:14
  • $\ln k$ is obviously positive because $k \ge 2$. – HorizonsMaths Mar 30 '14 at 14:55