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?
Question about estimating a summation
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$\begingroup$
sequences-and-series
analysis
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0compare to something like $\int_0^{2\pi}\int_t^{\infty}r^{-5}rdrd\theta=2\pi t^{-3}/3$ – 2011-12-13
1 Answers
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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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0Oh - a lattice now. Ok. – 2011-12-13