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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.

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  • $\begingroup$ This makes me think of Ramanujan's evaluation $\zeta(1)=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\int_1^n\frac1xdx\right)$ $\endgroup$ – Simply Beautiful Art Oct 9 '16 at 21:19
  • $\begingroup$ @Simple Art ....your result being equal to the Euler constant $\gamma=0.577\cdots$. $\endgroup$ – Jean Marie Oct 9 '16 at 21:34
  • $\begingroup$ @JeanMarie I mean, its a special value. But when you Ramanujan sum things like $1^p+2^p+3^p+\dots$, you get the Riemann zeta function (I think). $\endgroup$ – Simply Beautiful Art Oct 9 '16 at 21:40
  • $\begingroup$ @Simple Art I do agree that it is a special value, but sometimes, observing special values indicate some other tracks to follow. $\endgroup$ – Jean Marie Oct 9 '16 at 21:45
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    $\begingroup$ @SimpleArt Yes I know, but you wrote several times $\zeta(1)$ for $\displaystyle\overset{\mathfrak{R}}{\sum} n^{-1} = \gamma$, which is a nonsense so find another notation (in particular $\displaystyle\overset{\mathfrak{R}}{\sum} n^{-s}$ isn't continuous at $s=1$...). And as you probably know, the Ramanujan summation isn't a very common summation method, it is one of the most complicated, and many people use $\displaystyle\overset{\mathfrak{R}}{\sum} a_n = \alpha$ as a short-hand for "with some summation method -try them all- you should get $\sum a_n = \alpha$" $\endgroup$ – reuns Oct 10 '16 at 16:49
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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 $Re(s) > 0$ : $$\lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right) + \frac{N^{1-s}}{s-1} = \lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right) + \frac{(N+1)^{1-s}}{s-1} = F(s)+ \frac{1}{s-1}= \zeta(s)$$

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  • $\begingroup$ Oh, nice. I actually remember seeing this before done in reverse. Just couldn't remember where I saw it... +1 $\endgroup$ – Simply Beautiful Art Oct 10 '16 at 14:56
  • $\begingroup$ May I ponder to ask what happens for $-1<\Re(s)<0$ for $F(s)$? Seems convergence is possible in that region, but probably doesn't come out to give the Riemann zeta function(?) $\endgroup$ – Simply Beautiful Art Oct 10 '16 at 15:27
  • $\begingroup$ @SimpleArt apply the same trick again : $n^{-s} - \int_n^{n+1} x^{-s}dx - \frac{1}{2}\int_n^{n+1} s x^{-s-1}dx = \mathcal{O}(x^{-s-2})$. $\endgroup$ – reuns Oct 10 '16 at 15:33
  • $\begingroup$ I mean, how can $F(0)=\zeta(0)+1$? for example. If what you say is true, $F(0)$ is the limit point as $s\to0^-$, which should work out since $F$ is analytic, right? $\endgroup$ – Simply Beautiful Art Oct 10 '16 at 15:35
  • $\begingroup$ @SimpleArt $\lim_{s \to 0^+} F(0) = \zeta(0)+1$ yes $\endgroup$ – reuns Oct 10 '16 at 15:38
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty}\pars{\sum_{x = 1}^{n}x^{-1/3} - {3 \over 2}\,n^{2/3}} = \zeta\pars{1 \over 3}}$

\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.

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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.

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  • $\begingroup$ How does one calculate such a sum? $\endgroup$ – Ahmed S. Attaalla Oct 9 '16 at 22:24
  • $\begingroup$ @AhmedS.Attaalla Tad more specific please? $\endgroup$ – Simply Beautiful Art Oct 9 '16 at 22:25
  • $\begingroup$ @AhmedS.Attaalla I honestly don't know. I'm not very familiar with this method of summing divergent series. My naïve approach would be connecting zeta regularization to $\sum_{n\ge1}^\Re n^s$ somehow. $\endgroup$ – Simply Beautiful Art Oct 9 '16 at 22:30

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