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!
Decay for the tail of a series.
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sequences-and-series
asymptotics
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0You can easily derive one from the integral test. – 2012-07-27
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3Do you mean $k^{-p}$? As written the sum diverges. – 2012-07-27
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0@RossMillikan: Surely he does. It is interesting to note how easy it is to overlook such trivial mistakes. – 2012-07-27
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1@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. – 2012-07-27
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0The case where $p=2$ was examined in detail at this [MSE link](http://math.stackexchange.com/questions/685435/trying-to-get-a-bound-on-the-tail-of-the-series-for-zeta2). – 2014-03-27
1 Answers
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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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1For finer-grained estimates based on refinements of the above idea, please see the [Euler-Maclaurin](http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula) summation formula. – 2012-07-27