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Can anyone prove or reference a proof for the following bound (unless it's not true!)

$$\sum_{|\underline{k}|_{\infty} > M} \frac{1}{((k_1)^2 + (k_2)^2 )^2} \leq \frac{C}{M^2}$$

where $\underline{k} = (k_1,k_2)\in\mathbb{Z}^2$, $|\underline{k}|_{\infty} = \max(|k_1|,|k_2|)$, and $C$ is a constant independent of $M$. I'm trying to generalize to find a bound for

$$\sum_{|\underline{k}|_{\infty} > M} \left(\left(\frac{k_1}{a}\right)^2 + \left(\frac{k_2}{b}\right)^2 \right)^{-2} $$

where $a,b>0$. Obviously you can bound above by taking $d=\min(a,b)$ and then apply the first inequality to get $d^4 \frac{C}{M^2}$. But I don't want to use that, and I want something a bit sharper.

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  • $\begingroup$ $|\underline{k}|_{\infty} = \max(|k_1|,|k_2|)$, not $\max(k_1,k_2)$. $\endgroup$ – Olórin Dec 19 '14 at 14:32
  • $\begingroup$ sorry, editing now $\endgroup$ – Frubiclé Dec 19 '14 at 14:32
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We need only to bound the sum $$T(M)=\sum_{k_1\geq k_2\geq 1, k_1\geq M}\frac{1}{(k_1^2+k_2^2)^2}$$ We have: $$T(M)=\sum_{k_1\geq M}\frac{1}{k_1^3}A(k_1)$$ with $$A(k_1)=\frac{1}{k_1}\sum_{k_2=1}^{k_1}\frac{1}{(1+(k_2/k_1)^2)^2}$$

$A(k_1)$ is a Riemann sum, and has as limit $\displaystyle \int_0^1\frac{dx}{(1+x^2)^2}$. Hence $A(k_1)$ is bounded, say by $c$. Hence $$T(M)\leq c \sum_{k_1\geq M}\frac{1}{k_1^3}$$ and it is easy to finish.

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