By means of Ramanujan Summation, one can sum divergent values of the Riemann-zeta function. For instance, we have: $$ \zeta(-2k) = 1 + 2^{2k} + 3^{2k} + \dots = 0 ( \mathfrak{R} ) , $$ and $$ \zeta(-(2k - 1)) = 1 + 2^{2k-1} + 3^{2k-1} + \dots = - \frac{B_{2k} } {2k} (\mathfrak{R} ) . $$ Another interesting result is: $$ \sum_{n=1}^{\infty} \frac{1}{n} = \gamma (\mathfrak{R} ) .$$

From now on, I will slighty deviate from the wikipedia notation. I will write the Ramanujan summation symbol as an operator, rather than a strange-looking letter at the end of the equation. For instance, I will write: $R( \zeta(1))= \gamma$. This will make asking the questions more convenient.

Now, to the questions. I am wondering how to sum arithmetic combinations of these sums. Let $D_1 , D_2 , \dots , D_i, \dots , D_n $ be divergent series that are summable by means of the Ramanujan summation method. I am wondering how to sum the following:

  1. What is $R(D_1 + D_2)$ ? How does it relate to $R(D_1)$ and $R(D_2)$ ? In general: How does $R( \sum_{i \in I} D_i ) $ relate to $\cup_{i \in I} \{ R(D_i) \} $?
  2. What is $R(D_1 \cdot D_2)$? How does it relate to $R(D_1)$ and $R(D_2)$? Same questions apply to summing the product of $i \in I$ divergent series.

In case it is not possible to answer these questions in general: I am also interested in particular examples. For instance, I would also be glad to know what the result of $R(\zeta(-2) + \zeta(-3))$ or $R(\zeta(-3) + \zeta(-5))$ is. I am particularly interested in finding the value of $R( (\zeta(1))^3 ) $


Q1: The $R$ operator is linear, so $R(\sum_{i\in I} D_i)=\sum_{i\in I} R(D_i)$ at least for finite $I$.

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  • $\begingroup$ How do you know the $R$ operator is linear? $\endgroup$ – Max Muller Apr 30 '13 at 18:19
  • $\begingroup$ @Max: perhaps this small treatize on the representation of the Euler-MacLaurin-formula in terms of a matrix-operation might be interesting, because the Euler-MacLaurin-formula is at the heart of the Ramanujan-summation. (Having this in a matrix form one might say that the "Ramanujan-summation is a linear operator" .) see: go.helms-net.de/math/binomial_new/EulerMacLaurin.pdf $\endgroup$ – Gottfried Helms Aug 26 '14 at 23:11

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