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

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?

share|cite|improve this question
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]:… – 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


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


which after




share|cite|improve this answer

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

You will be able to prove

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

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 $m\geq 2$

\begin{equation} 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} \end{equation}

This would give you

\begin{equation} |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} \end{equation}

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

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

share|cite|improve this answer

You should note that the Cauchy principlal value of $\zeta(1)$ is $\gamma$:


Saying $\zeta(1)=\infty$ is wrong because zeta has no limit at that point (except for directional limits).

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.