*(This Question was taken from [MSE](https://math.stackexchange.com/questions/38731/values-of-the-riemann-zeta-function-and-the-ramanujan-summation-how-strong-is-t). As Eric Naslund pointed out there, this [question](https://mathoverflow.net/questions/3204/does-any-method-of-summing-divergent-series-work-on-the-harmonic-series) is relevant. The summation method mentioned in this question is actually a good answer to it.)*


The [Ramanujan Summation](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/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?



Values of the Riemann zeta function and the Ramanujan summation - How strong is the connection?
The answer can be obtained with the [following interpretation](http://en.wikipedia.org/wiki/Ramanujan_summation) 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](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/Hurwitz_zeta_function#Taylor_series) 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.


