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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]: http://mathoverflow.net/questions/3204/does-any-method-of-summing-divergent-series-work-on-the-harmonic-series2011-05-12
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    @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).2011-08-13

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