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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? – Henry Dec 12 '11 at 23:46
I consider $x$ to be a lattice point in the standard lattice. I edited the question slightly so that it should make sense now. – THK Dec 12 '11 at 23:53
what dimension is your "standard lattice?" – yoyo Dec 12 '11 at 23:57
If $x$ ranges over the integers, the sum is $0$. – Brian M. Scott Dec 12 '11 at 23:58
@yoyo: The lattice points should be in $\mathbb{R}^{2}$, I have edited the question to reflect this. Thanks! – THK Dec 13 '11 at 0:03
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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. – stopple Dec 13 '11 at 0:08
Oh - a lattice now. Ok. – mixedmath♦ Dec 13 '11 at 0:39

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