Let $h,m\in[1,\infty)$. I would like to verify that the following sum diverges logarithmically \begin{equation} \sum_{d=0}^{\infty} \frac{2d+1}{2h^2(1+\frac{d(d+1)}{h^2})(1+\frac{d(d+1)}{h^2m ^2})^{2}}\,. \end{equation} Roughly, it should be proportional to $\log(m^2)$. Since, we know that the Green function of Laplacian in $\mathbb{R}^2$ diverging logarithmically. I got this sum as the trace of resolvent of some regularization of Laplacian defined on two sphere of radius $h$. I wonder if I can make use of "Selberg’s Trace Formula"? I would be enormously pleased if someone could explain it to me. Many thanks.
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1$\begingroup$ Some general techniques: (a) try estimating a single dyadic block $\sum_{2^{j-1} \leq d < 2^j}$ first, then sum in $j$ later. (b) The point of working with a dyadic block is that one can greatly simplify algebraic expressions up to constants. For instance, if $2^{j-1} \leq d < 2^j$, then $d$ and $d+1$ are both comparable to $2^j$. (c) Up to constants, one can approximate the sum of two non-negative quantities $a+b$ by $a$ when $a \gtrsim b$ and $b$ when $b \gtrsim a$. This will help you simplify expressions such as $1 + d(d+1)/h^2$ after splitting into cases. $\endgroup$– Terry TaoCommented Sep 29, 2023 at 19:47
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$\begingroup$ @TerryTao Thank you very much. My idea was to write this sum as an integral using floor function, then split the integration between $0\leq d<1$ and $1\leq d<\infty$. The first interval causes no problem as it is converging but the second one is problematic. I will try your suggestion. Thank you once again. $\endgroup$– AzamCommented Sep 30, 2023 at 7:12
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$\begingroup$ Perhaps we do not need to use dyadic, but rather directly discuss cases such as d(d+1)<h, h<d(d+1)<mh, and d(d+1)>mh, and then sum up respectively. It should be noted that the contributions from the first and third parts are constants (dependent on h), while the contribution from the middle part is (\log m). $\endgroup$– HaggiCommented Apr 13 at 7:38
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