# Truncating $p$-series

Let $p \in \mathbb{R}$. Series of the form $$S_p = \sum_{k \in \mathbb{N^*}} \frac{1}{k^p}$$ converge if and only if $p > 1$. Let us define $$S_{p, n} \triangleq \sum_{k = 1}^{n} \frac{1}{k^p},$$ the truncation of the $p$-series at its $n$-th term. (It is evident that $S_{p, n} \space \xrightarrow{n\to\infty} \space S_p$.) Is there a closed expression for $S_{p, n}$, at least for some values of $p$ (integers must be easier)?

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\stackrel{n \to \infty}{\longrightarrow} should do the trick. – jd.r May 14 '11 at 0:41
$\overset{n\rightarrow \infty }{\longrightarrow }$ can be obtained with \overset{n\rightarrow \infty }{\longrightarrow } – Américo Tavares May 14 '11 at 0:54
I deleted an answer because I replied to a different question (closed form for $S_p$, with $p$ even or negative) – Américo Tavares May 14 '11 at 0:58
It depends on what you mean by "closed form expression." It is surprisingly difficult to give a precise definition of this. – Qiaochu Yuan May 14 '11 at 3:32
Here's a plausibility argument on why you're supposed to expect the polygamma functions (as shown in Robert's and Eric's answers) to pop up: recall the usual gamma function relation $\Gamma(n+1)=n\Gamma(n)$ and then derive from it $\log\Gamma(n+1)=\log\,n+\log\Gamma(n)$. As noted, the polygamma functions are derivatives of $\log\Gamma$, and through the identity $$\frac{\mathrm d^n}{\mathrm dz^n}\log\,z=\frac{(-1)^{n-1}(n-1)!}{z^n}$$, you'll find that a connection formula for $\psi^{(n)}(z+k)$ and $\psi^{(n)}(z)$ would necessarily involve a series of reciprocal powers. – J. M. May 14 '11 at 4:34

In Maple notation, for $p \ge 2$ integer, $$S_{p,n} = \zeta(p) + \frac{(-1)^{p-1}}{(p-1)!} \Psi(p-1,n+1),$$ where $\psi(m,\space·)$ is the $m$-th polygamma function.

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For completeness: if $p=1$, $$\sum_{k=1}^n k^{-1}=\gamma+\psi^{(0)}(n+1)$$ where $\gamma$ is the Euler-Mascheroni constant. (The zeta function diverges for $p=1$.) For noninteger $p$, there has been research into analytically continuing the polygamma function such that Robert's expression still makes sense. See this for instance. – J. M. May 14 '11 at 3:45
I could be wrong, but I doubt this is what Luke had in mind when he asked for a closed form. – Gerry Myerson May 14 '11 at 11:31

Yes, there is. These are called the generalized harmonic numbers.

With polygamma function: I suggest looking at this paper, they have many interesting things. In particular, if we define $$H_n(z;r)=\sum_{k=1}^n\frac{1}{(z+k-1)^r}$$ (even more generalized) then we have for integers $p$ $$H_n(z;p)=\frac{(-1)^{p-1}}{(p-1)!}(\psi^{(p-1)}(z+n)-\psi^{(p-1)}(z))$$ where $$\psi^{(k-1)}(z)=\frac{d^k}{dx^k}\left(\log\Gamma(x)\right) \biggr|_{x=z}.$$ This is equation (1.14) of the paper, but I think they forgot a factor of $(p-1)!$ on the denominator. Taking the case $z=1$ gives what you are looking for.

With zeta function: we can write $$H_n(r)=\sum_{k=1}^n\frac{1}{k^r}=\zeta(r)-\zeta(r,n+1),$$ where $\zeta(s,a)$ is the Hurwitz zeta function. This $\space$ really clear, but by using identities regarding $\zeta(s,a)$ we can change the identity into $\space$ previous one, and other things.

Hope that helps.

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In short, the (generalized) harmonic numbers and the polygamma function are essentially the same thing (differing only by a constant); and are tightly related to Hurwitz zeta. – J. M. May 14 '11 at 2:59
@J.M. That is a nice way to put it! – Eric Naslund May 14 '11 at 3:06
A tiny tip, BTW: for linking to SD, it usually makes better sense to use the DOI instead of the actual SD link; for some reason they tend to go stale quickly. In your case, http://dx.doi.org/10.1016/S0096-3003(01)00172-2 would be a better link to the paper in your answer. – J. M. May 14 '11 at 3:36
I could be wrong, but I doubt this is what Luke had in mind when he asked for a closed form. – Gerry Myerson May 14 '11 at 11:31
On the other hand, Luke did not seem to complain about Eric's and Robert's answers... – J. M. May 14 '11 at 11:38

To the best of my knowledge, there is no closed form expression for $S_{p,n}$.

EDIT: Let me amend my answer to read, to the best of my knowledge, there is no closed form expression in terms of functions familiar to, say, students of 1st-year Calculus.

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