Values of the Riemann zeta function and the Ramanujan summation - How strong is the connection? (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?
 A: 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.
