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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.

Ahmed S. Attaalla
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  • This makes me think of Ramanujan's evaluation $\zeta(1)=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\int_1^n\frac1xdx\right)$ – Simply Beautiful Art Oct 09 '16 at 21:19
  • @Simple Art ....your result being equal to the Euler constant $\gamma=0.577\cdots$. – Jean Marie Oct 09 '16 at 21:34
  • @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). – Simply Beautiful Art Oct 09 '16 at 21:40
  • @Simple Art I do agree that it is a special value, but sometimes, observing special values indicate some other tracks to follow. – Jean Marie Oct 09 '16 at 21:45
  • @JeanMarie Good advice. :) But bookmark those special values before leaving the track. – Simply Beautiful Art Oct 09 '16 at 21:46
  • @user1952009 That is not what I meant. I was speaking of the Ramanujan evaluation of $\zeta(1)$, not $\zeta(1)$ itself. – Simply Beautiful Art Oct 10 '16 at 15:15
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    @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 $\sum a_n = \alpha$" – reuns Oct 10 '16 at 16:49
  • See for example https://hal-unice.archives-ouvertes.fr/hal-01150208/document using $\displaystyle\overset{\mathfrak{R}}{\sum}$ is often a pitty – reuns Oct 10 '16 at 16:53
  • @user1952009 yes, I can agree with that, but it's not like I didn't specify here. – Simply Beautiful Art Oct 10 '16 at 20:27
  • @SimpleArt well I showed you two problems : forget about $\zeta(1) = \gamma$, and try avoiding $\displaystyle\overset{\mathfrak{R}}\sum$ as possible – reuns Oct 10 '16 at 20:33

3 Answers3

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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 $Re(s) > 0$ : $$\lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right) + \frac{N^{1-s}}{s-1} = \lim_{N \to \infty} \left(\sum_{n=1}^N n^{-s}\right) + \frac{(N+1)^{1-s}}{s-1} = F(s)+ \frac{1}{s-1}= \zeta(s)$$

reuns
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  • Oh, nice. I actually remember seeing this before done in reverse. Just couldn't remember where I saw it... +1 – Simply Beautiful Art Oct 10 '16 at 14:56
  • 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(?) – Simply Beautiful Art Oct 10 '16 at 15:27
  • @SimpleArt apply the same trick again : $n^{-s} - \int_n^{n+1} x^{-s}dx - \frac{1}{2}\int_n^{n+1} s x^{-s-1}dx = \mathcal{O}(x^{-s-2})$. – reuns Oct 10 '16 at 15:33
  • 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? – Simply Beautiful Art Oct 10 '16 at 15:35
  • @SimpleArt $\lim_{s \to 0^+} F(0) = \zeta(0)+1$ yes – reuns Oct 10 '16 at 15:38
  • $F(0)=0$ from the original definition of $F$, whereas $\zeta(0)+1=1/2$. – Simply Beautiful Art Oct 10 '16 at 15:38
  • @SimpleArt well $\lim_{s \to 0^+} F(s) \ne 0$ then – reuns Oct 10 '16 at 15:42
  • But if $F$ is analytic, then $\lim_{s\to0^+}F(s)$ should equal $F(0)$? Huh. I want to actually see some numerical values for the limit now. – Simply Beautiful Art Oct 10 '16 at 15:43
  • @SimpleArt yes try. And the expression for $F(s)$ is analytic on $Re(s) > 0$. – reuns Oct 10 '16 at 15:44
  • Was thinking that if the representation is analytic on $\Re(s)>0$ but still converges for other values, then it should be analytic there too. :/ Will do when I get home. – Simply Beautiful Art Oct 10 '16 at 15:45
  • @SimpleArt For example $h(s) = \sum_{n \ge 1} (-1)^{n+1} n^{-s}$ is analytic on $Re(s) > 1$ whatever the order of summation, but (as we said yesterday) changing the order of summation you can get the value you want for this series at $s= 1$, for example $h(1) = 0$, whereas $\lim_{s \to 1^+} h(s) = \ln 2$ – reuns Oct 10 '16 at 15:46
  • But that is due to $h(1)$ being conditionally convergent. Here, $F(s)$ is not conditionally convergent at $s=0$. – Simply Beautiful Art Oct 10 '16 at 20:25
  • @SimpleArt you don't get it. Let $g(s) = s \sum_{n=1}^\infty n^{-s-1}$. Then $\lim_{s \to 0} g(s) = \lim_{s \to 1} (s-1)\zeta(s) = 1$, while under your reasoning $g(0) = 0$. Here it is exactly the same, since $n^{-s} - \int_n^{n+1} x^{-s}dx= \frac{s}{2} n^{-s-1} + \mathcal{O}(n^{-s-2})$ – reuns Oct 10 '16 at 20:29
  • @SimpleArt yes a typo $\lim_{s \to 0}$ – reuns Oct 10 '16 at 20:34
  • Hm, I understand what you're getting at, the sum being a sorts of double limit. But I still don't feel the same about the $F(0)$ case, it just doesn't... feel the same. – Simply Beautiful Art Oct 10 '16 at 20:39
  • That representation implies $F(0)$ diverges to infinity btw. – Simply Beautiful Art Oct 10 '16 at 20:42
  • @SimpleArt I swear it is exactly the same, since I proved on the 1st line that $F(s) = \frac{1}{2}\sum_{n=1}^\infty n^{-s-1}+ \mathcal{O}(\sum_{n=1}^\infty |n^{-s-2}|)$ for $Re(s) > 0$ (well I didn't write it but $\int_n^{n+1} \int_n^x s t^{s-1}dt dx = \int_n^{n+1} \int_n^x s (n^{s-1}+\mathcal{O}(n^{-s-2}))dt dx = s n^{-s-1}+\mathcal{O}(n^{-s-2})$ – reuns Oct 10 '16 at 20:43
  • @SimpleArt no it doesn't diverge, because $\frac{s}{2} n^{-s-1} \to 0$ as $s \to 0$ – reuns Oct 10 '16 at 20:43
  • But $\sum_{n=1}^\infty n^{-s-1}+\mathcal O(\sum_{n=1}^\infty|n^{-s-2}|)\to\infty$ as $s\to0$. – Simply Beautiful Art Oct 10 '16 at 20:45
  • @SimpleArt Yes but It is a typo, I meant $F(s) = \frac{1}{2} \sum_{n=1}^\infty s n^{-s-1} + \mathcal{O}(\sum_{n=1}^\infty |n^{-s-2}|)$ ... Come on, did you read my answer ? – reuns Oct 10 '16 at 20:52
  • Um, but that wasn't in your answer?! *facepalm* – Simply Beautiful Art Oct 10 '16 at 20:54
  • @SimpleArt so now everything is clear, the $1/2$ that disappeared in $\lim_{s \to 0} F(s) = \zeta(0)+1 = 1/2$ while $F(0) = 0$ is due to the $\frac{1}{2} \sum_{n=1}^\infty s n^{-s-1}$ term that is not continuous, with its limit being $1/2$ while its value at $0$ is $0$ – reuns Oct 10 '16 at 20:56
  • @SimpleArt There is the analytic continuation you always dreamed about

    $\zeta(s) = \frac{1}{s-1}+\sum_{n=1}^\infty n^{-s}-\int_n^{n+1} x^{-s}dx =\frac{1}{s-1}+\sum_{n=1}^\infty \int_n^{n+1}\int_n^x s t^{-s-1}dtdx$

    $ = \frac{1}{s-1}+\frac{s}{2}\zeta(s+1)+\sum_{n=1}^\infty \int_n^{n+1}\int_n^x s (t^{-s-1}-n^{-s-1})dtdx$

    $= \frac{1}{s-1}+\frac{s}{2}\zeta(s+1)+\sum_{n=1}^\infty \int_n^{n+1}\int_n^x s \int_n^t (s+1)u^{-s-2}dudtdx$

    $ = \frac{1}{s-1}+\frac{s}{2}\zeta(s+1)+\frac{s(s+1)}{2.3}\zeta(s+2)+\sum_{n=1}^\infty \int_n^{n+1}\int_n^x s \int_n^t (s+1)(u^{-s-2}-n^{-s-2})dudtdx$

     – reuns Oct 10 '16 at 21:15
  • $= \frac{1}{s-1}+\frac{s}{2}\zeta(s+1)+\frac{s(s+1)}{2.3}\zeta(s+2)+\sum_{n=1}^\infty \int_n^{n+1}\int_n^x s \int_n^t (s+1)\int_n^u (s+2)v^{-s-3}dvdudtdx$

    $ =\frac{1}{s-1}+\frac{s}{2}\zeta(s+1)+\frac{s(s+1)}{2.3}\zeta(s+2)+\frac{s(s+1)}{2.3.4}\zeta(s+3)+\sum_{n=1}^\infty \int_n^{n+1}\int_n^x s \int_n^t (s+1)\int_n^u (s+2)(n^{-s-3}-v^{-s-3})dvdudtdx$

     – reuns Oct 10 '16 at 21:15
  • Lol, I get it. Interesting expansion. By the way, you can use \. to create spaces between all the d#'s. – Simply Beautiful Art Oct 10 '16 at 21:22
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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.

Felix Marin
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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.

Simply Beautiful Art
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