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.

• 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 9 '16 at 21:19
• @Simple Art ....your result being equal to the Euler constant $\gamma=0.577\cdots$. – Jean Marie Oct 9 '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 9 '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 9 '16 at 21:45
• @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

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

• 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


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

\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.

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.

• How does one calculate such a sum? – Ahmed S. Attaalla Oct 9 '16 at 22:24
• @AhmedS.Attaalla Tad more specific please? – Simply Beautiful Art Oct 9 '16 at 22:25
• @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. – Simply Beautiful Art Oct 9 '16 at 22:30