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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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    @Theo: Thanks for the explanation.2011-05-14

3 Answers 3

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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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    I could be wrong, but I doubt this is what Luke had in mind when he asked for a closed form.2011-05-14
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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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    @Gerry @J. M. This is closed enough for me. It's neither recursive nor kept on being a truncated sum. I voted it up.2011-05-14
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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.