Limit related to $\zeta(x)$ I'm noticing some things:
$$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/3}-\frac{3}{2}n^{2/3} \right)=\zeta(1/3)$$
Note $\int n^{-1/3} dn=\frac{3}{2}n^{2/3}+c$
$$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/2}-\frac{2}{1}n^{1/2} \right)=\zeta(1/2)$$
And
$$\int n^{-1/2}dn=2n^{1/2}+c$$
It seems as though
$$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/s}-\frac{s}{s-1}n^{(s-1)/s} \right)=\zeta(1/s)$$
If $s \neq 1$, may someone please explain why.
 A: As $n \to \infty$ : $$n^{-s} -\int_n^{n+1} x^{-s}dx = \int_n^{n+1} (n^{-s}-x^{-s})dx = \int_n^{n+1} \int_n^x s t^{-s-1}dt dx = \mathcal{O}(n^{-s-1})$$
Thus
$$F(s) = \sum_{n=1}^\infty \left( n^{-s}-\int_n^{n+1} x^{-s}dx\right) = \lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right)-\int_1^{N+1} x^{-s}dx$$ $$ = \lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right) - \frac{1-(N+1)^{1-s}}{s-1}$$
converges and is analytic for $Re(s) > 0$.
But for $Re(s) > 1$, $\lim_{N \to \infty} (N+1)^{1-s} = 0$ so that 
$$F(s) = \frac{-1}{s-1}+\sum_{n=1}^\infty n^{-s} = \frac{-1}{s-1}+\zeta(s)$$
And by analytic continuation this stays true for $Re(s) > 0$ (or if you prefer by the identity theorem for complex analytic functions).
Finally, since for $Re(s) > 0$  : $\lim_{N \to \infty} (N+1)^{1-s}-N^{1-s} = 0$, you get that for $Re(s) > 0$ : $$\lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right) + \frac{N^{1-s}}{s-1} = \lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right) + \frac{(N+1)^{1-s}}{s-1} = F(s)+ \frac{1}{s-1}= \zeta(s)$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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$\ds{\lim_{n \to \infty}\pars{\sum_{x = 1}^{n}x^{-1/3} - {3 \over 2}\,n^{2/3}} =
\zeta\pars{1 \over 3}}$

\begin{align}
\sum_{x = 1}^{n}x^{-1/3} & =
{3 \over 2}\sum_{x = 1}^{n}{x - 1/3 \over x^{1/3}} -
{3 \over 2}\sum_{x = 1}^{n}x^{2/3}
+ {3 \over 2}\sum_{x = 0}^{n - 1}{1 \over \pars{x + 1}^{1/3}}
\\[5mm] & =
{3 \over 2}\sum_{x = 1}^{n}{x - 1/3 \over x^{1/3}} -
{3 \over 2}\sum_{x = 1}^{n}x^{2/3}
+ {3 \over 2} + {3 \over 2}\sum_{x = 1}^{n}{1 \over \pars{x + 1}^{1/3}} -
{3 \over 2}{1 \over \pars{n + 1}^{1/3}}
\\[1cm] & =
-\,{3 \over 2}\sum_{x = 1}^{n}
\bracks{{x \over \pars{x + 1}^{1/3}} - {x - 1/3 \over x^{1/3}}} + {3 \over 2} +
{3 \over 2}\ \underbrace{%
\bracks{\sum_{x = 1}^{n}\pars{x + 1}^{2/3} - \sum_{x = 1}^{n}x^{2/3}}}
_{\ds{-1 + \pars{n + 1}^{2/3}}}
\\ & \phantom{=\,\,}-\,{3 \over 2}{1 \over \pars{n + 1}^{1/3}}
\end{align}

\begin{align}
&\mbox{Then,}\quad\lim_{n \to \infty}\pars{%
\sum_{x = 1}^{n}x^{-1/3} - {3 \over 2}\,n^{2/3}}\ =\
\overbrace{-\,{3 \over 2}
\sum_{x = 1}^{\infty}
\bracks{{x \over \pars{x + 1}^{1/3}} - {x - 1/3 \over x^{1/3}}}}
^{\ds{=\ \zeta\pars{1 \over 3}}}\label{1}\tag{1}
\\[5mm] & \phantom{=}+\
\underbrace{{3 \over 2}\lim_{n \to \infty}\bracks{\pars{n + 1}^{2/3} -
\,{3 \over 2}{1 \over \pars{n + 1}^{1/3}} - n^{2/3}}}_{\ds{=\ 0}}\ =\
\bbox[10px,#ffe,border:1px dotted navy]{\ds{\zeta\pars{1 \over 3}}}
\end{align}


The series, in the \eqref{1} RHS, is a $\ds{\zeta}$-representation which is obtained by a rearrange of the 'original definition' such that it extends the series validity range. Details are given in the above cited link.

A: From the Wikipedia of Ramanujan summation (a few paragraphs down):

It has been proposed to use of $C(1)$ rather than $C(0)$ as the result of Ramanujan's summation, since then it can be assured that one series 
  $\sum _{k=1}^{\infty }f(k)$ 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 $\int_{1}^{2}R(t)dt=0$. This definition of Ramanujan's summation (denoted as $\sum _{n\geq 1}^{\Re }f(n)$) does not coincide with the earlier defined Ramanujan's summation, $C(0)$, nor with the summation of convergent series, but it has interesting properties, such as: If $R(x)$ tends to a finite limit when $x\to1^+$, then the series 
  $\sum _{n\geq 1}^{\Re }f(n)$ is convergent, and we have
$$\sum_{n\ge1}^\Re f(n)=\lim_{N\to\infty}\left(\sum_{n=1}^Nf(n)-\int_1^Nf(t)dt\right)$$

which seems to be exactly what you are looking for.  More context is provided in the link.
