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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.

Aban
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    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. – Terry Tao Sep 29 '23 at 19:47
  • @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. – Aban Sep 30 '23 at 07:12

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