# Values of the Riemann Zeta function and the Ramanujan Summation - How strong is the connection?

The Ramanujan Summation of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, $$\zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 (\mathfrak{R})$$ (for non-negative integer $k$) and $$\zeta(-(2n+1))=-\frac{B_{2k}}{2k} (\mathfrak{R})$$ (again, $k \in \mathbb{N}$). Here, $B_k$ is the $k$'th Bernoulli number. However, it does not hold when, for example, $$\sum_{n=1}^{\infty} \frac{1}{n}=\gamma (\mathfrak{R})$$ (here $\gamma$ denotes the Euler-Mascheroni Constant) as it is not equal to $$\zeta(1)=\infty$$.

Question: Are the first two examples I stated the only instances in which the Ramanujan summation of some infinite series coincides with the values of the Riemann zeta function?

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This [math overflow question][1] will certainly be of interest. It is regarding assigning a value to the divergent series $\zeta(1)$, as the harmonic series seems to be hard to assign such a value to. [1]: mathoverflow.net/questions/3204/… –  Eric Naslund May 12 '11 at 17:39
@Eric: Comments don't allow biblio-style hyperlinkes (the kind you get from clicking the link button in the graphical answer editor). You need an inline link, and the markup is [text](http://url). OP, I think the answer to your question may lie in what exactly the overlap between the Ramanujan and zeta summation methods is (family of divergent series that they agree on). –  anon Aug 13 '11 at 1:11

Ramanujan summation arises out of Euler-Maclaurin summation formula. Ramanujan summation is just (C, 1) summation. (See Cesàro summation)

You can find out easily from Euler-Maclaurin that

$$\sum_{k=1}^{\infty}\frac{1}{k}$$

is not (C, 1) summable.

Follow the method of Ramanujan below (which you can easily follow):

Using Euler-Summation we have

\begin{align*} \zeta(s) & = \frac{1}{s-1}+\frac{1}{2}+\sum_{r=2}^{q}\frac{B_r}{r!}(s)(s+1)\cdots(s+r-2) \\ & \phantom{=} -\frac{(s)(s+1)\cdots(s+q-1)}{q!}\int_{1}^{\infty}B_{q}(x-[x])x^{-s-q} ~dx \end{align*}

$\zeta(s)$ is the Riemann zeta function (Note $s=1$ is pole) . Note that right side has values even for $Re(s)<1$.

For example, putting $s=0$ we get $$\zeta(0)=-\frac{1}{2}.$$

If we put $s=-n$ (n being a positive integer) and $q=n+1$, we see the remainder vanishes and have

$$(n+1)\zeta(-n)=-1+\frac{n+1}{2}+\sum_{r=2}^{n+1}\frac{B_r(-1)^{r-1}}{r!}\binom{n+1}{r}$$

which after

$$\sum_{j=0}^{r}\binom{r+1}{j}B_j=0$$

gives

$$\zeta(-n)=-\frac{B_{n+1}}{n+1}.$$

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I would urge you to do analyze the harmonic series using Euler-Maclaurin Summation

You will be able to prove

$$\sum_{k\leq x}\frac{1}{k}=\log x+\gamma+O\left(\frac{1}{x}\right)$$

where $\gamma$ is the Euler-Mascheroni constant and $O(f)$ is big oh notation.

You just need to analyze the remainder term in Euler-Maclaurin summation using Fourier series of periodic Bernoulli polynomials. That is, for $m\geq 2$

$$B_m(x-[x])=-\frac{m!}{(2\pi i)^m}\sum_{n=-\infty,n\neq 0}^{n=\infty}\frac{e^{2\pi i nx}}{n^m}$$

This would give you

$$|B_{m}(x-[x])|\leq 2\frac{m!}{(2\pi)^m}\sum_{n=1}^{\infty}\frac{1}{n^m}\leq 2\frac{m!}{(2\pi)^m}\sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2\frac{m!}{3(2\pi)^m}$$

which is just perfect for estimating remainder in Euler-Maclaurin formula.

You can also try to prove Stirling's approximation using this method that is

$$n! = \sqrt{2 \pi n}~{\left( \frac{n}{e} \right)}^n \left( 1 + O \left( \frac{1}{n} \right) \right).$$

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