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Let $p>1$. I would like to have an estimate for the decay of the sequence $s_{n}=\sum_{k=n}^{\infty}k^{-p}$. Does anyone know of a bound of this type in the literature? Thanks!

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You can easily derive one from the integral test. – Harald Hanche-Olsen Jul 27 '12 at 15:03
Do you mean $k^{-p}$? As written the sum diverges. – Ross Millikan Jul 27 '12 at 15:04
@RossMillikan: Surely he does. It is interesting to note how easy it is to overlook such trivial mistakes. – Harald Hanche-Olsen Jul 27 '12 at 15:06
@HaraldHanche-Olsen: We have had interest recently in expressions like $\sum i = -\frac 1{12}$, so I was checking. I was writing up the same answer as Tom Cooney when I noticed it. I make these mistakes, too. – Ross Millikan Jul 27 '12 at 15:17
The case where $p=2$ was examined in detail at this MSE link. – Marko Riedel Mar 27 '14 at 0:20
up vote 4 down vote accepted

Look at the proof of the integral test of convergence for a sequence; we identify $s_n$ as upper and lower Riemann sums of integrals to get the bounds: $$ \int_{n+1}^\infty x^{-p} \ dx \leq \sum_{k=n}^\infty \frac{1}{k^p} \leq \int_{n}^\infty x^{-p} \ dx. $$ Evaluating the integrals, we then have $$ \frac{1}{p-1} \frac{1}{(n+1)^{p}} \leq \sum_{k=n}^\infty \frac{1}{k^p} \leq \frac{1}{p-1} \frac{1}{n^p}. $$

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For finer-grained estimates based on refinements of the above idea, please see the Euler-Maclaurin summation formula. – André Nicolas Jul 27 '12 at 15:57

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