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Let $t$ be a positive real number, with $x$ running over the standard lattice points in $\mathbb{R}^{2}$, is it true that $\sum_{|x| > t} t^{-5} = O(t^{-3})$? If so why?

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    Is $x$ an integer? A positive integer?2011-12-12
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    I consider $x$ to be a lattice point in the standard lattice. I edited the question slightly so that it should make sense now.2011-12-12
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    what dimension is your "standard lattice?"2011-12-12
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    If $x$ ranges over the integers, the sum is $0$.2011-12-12
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    @yoyo: The lattice points should be in $\mathbb{R}^{2}$, I have edited the question to reflect this. Thanks!2011-12-13
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    compare to something like $\int_0^{2\pi}\int_t^{\infty}r^{-5}rdrd\theta=2\pi t^{-3}/3$2011-12-13

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If we suppose that $t$ is bigger than 1, say, then we can approximate half of our sum with the integral $\displaystyle \int_t^\infty \frac{1}{x^5} = \frac{1}{4 t^4}$, and $\displaystyle\frac{1}{t^4} \in O\left(\frac{1}{t^3}\right)$.

EDIT (now that I know it's over a lattice)

Do what I did for one dimension above, but in two dimensions. $\displaystyle \int_0^{2 \pi} \int_t^\infty \frac{1}{r^5} r dr d\theta$ will do if you like polar coordinates.

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    The sum is over a lattice in $\mathbb R^2$, so you need a double integral to approximate it.2011-12-13
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    Oh - a lattice now. Ok.2011-12-13