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.

Solution 1:

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

Solution 2:

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

\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}

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\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.