How to bound this series

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I am currently reading through Tsybakov's Introduction to Nonparametric Statistics, and in it, he uses the estimate
$$left(sum_m=n^infty (m-1)^-2betaright)^1/2 = Oleft(n^-beta + 1/2right)$$
(here $beta > 0$ if that is important) or in other words,
$$sum_m=n^infty (m-1)^-2beta = Oleft(n^-2beta + 1right)$$
o more abstractly, one might even prove that (with proper assumptions on $k$ if necessary)
$$sum_m=n^infty (m-1)^-k = Oleft(n^-k + 1right)$$
This is a bit different than what I usually encounter, the trickiness arising because the summation index is in the base, not the exponent. How does one go about proving such an estimate?




sequences-and-series analysis




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  • 1




    Try a comparison series/integral...
    – Clement C.
    Jul 30 at 1:52










  • Right, that was silly...
    – stats_model
    Jul 30 at 1:58










  • We all fail to see something now and then... not that silly.
    – Clement C.
    Jul 30 at 1:59














up vote
0
down vote

favorite












I am currently reading through Tsybakov's Introduction to Nonparametric Statistics, and in it, he uses the estimate
$$left(sum_m=n^infty (m-1)^-2betaright)^1/2 = Oleft(n^-beta + 1/2right)$$
(here $beta > 0$ if that is important) or in other words,
$$sum_m=n^infty (m-1)^-2beta = Oleft(n^-2beta + 1right)$$
o more abstractly, one might even prove that (with proper assumptions on $k$ if necessary)
$$sum_m=n^infty (m-1)^-k = Oleft(n^-k + 1right)$$
This is a bit different than what I usually encounter, the trickiness arising because the summation index is in the base, not the exponent. How does one go about proving such an estimate?




sequences-and-series analysis




share|cite|improve this question















  • 1




    Try a comparison series/integral...
    – Clement C.
    Jul 30 at 1:52










  • Right, that was silly...
    – stats_model
    Jul 30 at 1:58










  • We all fail to see something now and then... not that silly.
    – Clement C.
    Jul 30 at 1:59












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am currently reading through Tsybakov's Introduction to Nonparametric Statistics, and in it, he uses the estimate
$$left(sum_m=n^infty (m-1)^-2betaright)^1/2 = Oleft(n^-beta + 1/2right)$$
(here $beta > 0$ if that is important) or in other words,
$$sum_m=n^infty (m-1)^-2beta = Oleft(n^-2beta + 1right)$$
o more abstractly, one might even prove that (with proper assumptions on $k$ if necessary)
$$sum_m=n^infty (m-1)^-k = Oleft(n^-k + 1right)$$
This is a bit different than what I usually encounter, the trickiness arising because the summation index is in the base, not the exponent. How does one go about proving such an estimate?




sequences-and-series analysis




share|cite|improve this question











I am currently reading through Tsybakov's Introduction to Nonparametric Statistics, and in it, he uses the estimate
$$left(sum_m=n^infty (m-1)^-2betaright)^1/2 = Oleft(n^-beta + 1/2right)$$
(here $beta > 0$ if that is important) or in other words,
$$sum_m=n^infty (m-1)^-2beta = Oleft(n^-2beta + 1right)$$
o more abstractly, one might even prove that (with proper assumptions on $k$ if necessary)
$$sum_m=n^infty (m-1)^-k = Oleft(n^-k + 1right)$$
This is a bit different than what I usually encounter, the trickiness arising because the summation index is in the base, not the exponent. How does one go about proving such an estimate?





sequences-and-series analysis





share|cite|improve this question










share|cite|improve this question




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asked Jul 30 at 1:41









stats_model

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55339







  • 1




    Try a comparison series/integral...
    – Clement C.
    Jul 30 at 1:52










  • Right, that was silly...
    – stats_model
    Jul 30 at 1:58










  • We all fail to see something now and then... not that silly.
    – Clement C.
    Jul 30 at 1:59












  • 1




    Try a comparison series/integral...
    – Clement C.
    Jul 30 at 1:52










  • Right, that was silly...
    – stats_model
    Jul 30 at 1:58










  • We all fail to see something now and then... not that silly.
    – Clement C.
    Jul 30 at 1:59







1




1




Try a comparison series/integral...
– Clement C.
Jul 30 at 1:52




Try a comparison series/integral...
– Clement C.
Jul 30 at 1:52












Right, that was silly...
– stats_model
Jul 30 at 1:58




Right, that was silly...
– stats_model
Jul 30 at 1:58












We all fail to see something now and then... not that silly.
– Clement C.
Jul 30 at 1:59




We all fail to see something now and then... not that silly.
– Clement C.
Jul 30 at 1:59















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