Values of the Riemann Zeta function and the Ramanujan Summation - How strong is the connection? 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: Ramanujan summation arises out of Euler-Maclaurin summation formula. Ramanujan summation is just (C, 1) summation. (See Cesàro summation)
You can find out easily from Euler-Maclaurin that
$$\sum_{k=1}^{\infty}\frac{1}{k}$$
is not (C, 1) summable.
Follow the method of Ramanujan below (which you can easily follow):
Using Euler-Summation we have
\begin{align*}
\zeta(s) & = \frac{1}{s-1}+\frac{1}{2}+\sum_{r=2}^{q}\frac{B_r}{r!}(s)(s+1)\cdots(s+r-2) \\
 & \phantom{=} -\frac{(s)(s+1)\cdots(s+q-1)}{q!}\int_{1}^{\infty}B_{q}(x-[x])x^{-s-q} ~dx
\end{align*}
$\zeta(s)$ is the Riemann zeta function (Note $s=1$ is pole) . Note that right side has values even for $Re(s)<1$.
For example, putting $s=0$ we get
$$\zeta(0)=-\frac{1}{2}.$$
If we put $s=-n$ (n being a positive integer) and $q=n+1$, we see the remainder vanishes and have
$$(n+1)\zeta(-n)=-1+\frac{n+1}{2}+\sum_{r=2}^{n+1}\frac{B_r(-1)^{r-1}}{r!}\binom{n+1}{r}$$
which after
$$\sum_{j=0}^{r}\binom{r+1}{j}B_j=0$$
gives
$$\zeta(-n)=-\frac{B_{n+1}}{n+1}.$$
A: I would urge you to do analyze the harmonic series using Euler-Maclaurin Summation 
You will be able to prove
\begin{equation}
\sum_{k\leq x}\frac{1}{k}=\log x+\gamma+O\left(\frac{1}{x}\right)
\end{equation}
where $\gamma$ is the Euler-Mascheroni constant and $O(f)$ is big oh notation.
You just need to analyze the remainder term in Euler-Maclaurin summation using Fourier series of periodic Bernoulli polynomials. That is, for $m\geq 2$
\begin{equation}
 B_m(x-[x])=-\frac{m!}{(2\pi i)^m}\sum_{n=-\infty,n\neq 0}^{n=\infty}\frac{e^{2\pi i nx}}{n^m} 
\end{equation}
This would give you
\begin{equation}
|B_{m}(x-[x])|\leq 2\frac{m!}{(2\pi)^m}\sum_{n=1}^{\infty}\frac{1}{n^m}\leq 2\frac{m!}{(2\pi)^m}\sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2\frac{m!}{3(2\pi)^m}
\end{equation}
which is just perfect for estimating remainder in Euler-Maclaurin formula. 
You can also try to prove Stirling's approximation using this method that is
\begin{equation}
n! = \sqrt{2 \pi n}~{\left( \frac{n}{e} \right)}^n \left( 1 + O \left( \frac{1}{n} \right) \right). 
\end{equation}
A: You should note that the Cauchy principlal value of $\zeta(1)$ is $\gamma$:
$$\lim_{h\to0}\frac{\zeta(1+h)+\zeta(1-h)}2=\gamma$$
Saying $\zeta(1)=\infty$ is wrong because zeta has no limit at that point (except for directional limits).
