(This Question was taken from MSE. As Eric Naslund pointed out there, this question is relevant. The summation method mentioned in this question is actually a good answer to it.)

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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    $\begingroup$ Actually $\gamma$ can be viewed as a regularized value of $\zeta(1)$ so this seems to me rather a case of "correct" evaluation. Since Ramanujan summation deals with divergent series, perhaps it is more appropriate to regard resuts of such summation as regularized values in some sense. $\endgroup$
    – Andrew
    May 13, 2011 at 15:33
  • $\begingroup$ @andrew I agree, I think we should view the Ramanujan summation as some kind of regularization. I wonder if zeta regularization and Ramanujan summation always coincide. If they don't: where and why not? Coming back to this question, I am especially interested in the case of $Re(z)=\frac{1}{2}$. I suppose that, If the Ramanujan summation method could sum the values on this vertical line, perhaps it could aid in determining the exact form of the zeros of the zeta function. $\endgroup$
    – Max Muller
    May 13, 2011 at 15:57
  • $\begingroup$ By the way, even if the Ramanujan summation wouldn't be able to sum these values, I think it would still be interesting to look for summation methods that could succeed in doing so. $\endgroup$
    – Max Muller
    May 13, 2011 at 16:01
  • $\begingroup$ @Max Muller: I like this question. What happens if you try $\zeta(1/2)$? $\endgroup$ Jul 13, 2011 at 17:00
  • $\begingroup$ @Eric: Thanks. (sorry for taking so long to respond.) I guess it amounts to plugging in (1/2) in the last formula provided by andrew (below). This would mean that $ \zeta(1/2) = 0.458780441... (\mathfrak{R}) $. $\endgroup$
    – Max Muller
    Aug 19, 2011 at 21:37

1 Answer 1


The answer can be obtained with the following interpretation of the Ramanujan summation:

More recently, the use of $C(1)$ has been proposed as Ramanujan's summation, since then it can be assured that one series admits one and only one Ramanujan's summation, defined as the value in 1 of the only solution of the difference equation $R(x) − R(x + 1) = f(x)$ that verifies the condition $\int_1^2 R(x)dx=0$.

The function $R(x)=\zeta(z,x)+C\;$ satisfy $R(x) − R(x + 1) = x^{-z}$, $x>0$, $z\in \mathbb C$, $z\ne-1$, where $$ \zeta(z,x)=\sum_{k=0}^\infty\frac1{(k+x)^z} $$ is the Hurwitz zeta function (or its analytic continuation for $\Re z\le 1$.) The value of $C\;$ can be found with the help of the shift formula for the derivative $\frac{\partial}{\partial x} \zeta(z,x)=-z \zeta(z+1,x)\;$: $$ \int_1^2 R(x)dx= \int_1^2(\zeta(z,x)+C)dx= \int_1^2 \left( -\frac1{z-1}\frac{\partial}{\partial x} \zeta(z-1,x)+C\right)dx= $$ $$ C-\frac1{z-1}(\zeta(z-1,2)-\zeta(z-1,1))=C+\frac1{z-1}=0.$$ Hence $C=-\frac1{z-1}$. Also $\zeta(z,1)=\zeta(z)$ and we have $$ \sum_{n=1}^\infty n^{-z}=\zeta(z)-\frac1{z-1}(\mathfrak{R}),\quad z\ne1. $$ So Ramanujan summation transforms the Riemann zeta function into the regularized zeta function. It explains why the value $\gamma$ should be expected for the summation of the harmonic series.


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