Absolute convergence of $\sum_{n,m\geq 1}\frac{1}{(n+im)^3}$ related to the sum of squares function As stated in the title, the series $\sum_{n,m\geq 1}\frac{1}{(n+im)^3}$ is absolutely convergent, if and only if
$$
\sum_{n,m\geq 1}\frac{1}{(n^2+m^2)^{3/2}}=\sum_{N\geq 2}\frac{r_2(N)}{N^{3/2}}<\infty\,,
$$
where $r_2(N)=|\{(n,m)\in\mathbb{N}^2\mid n^2+m^2=N\}|$ is a modification of the sum of squares function.
We have a trivial bound $r_2(N)<\sqrt{N}$ which is not sufficient. We rather need a bound $r_2(N)\leq N^{1/2-\epsilon}$ for some $\epsilon>0$.
Can you give a proof along these lines by bounding $r_2(N)$? I want an elementary proof without integration or anything beyond series and sequences, as this is exercise Number 16 in Königsbergers book Analysis 1 in the very beginning of the book (Chapter 6).
I browsed the questions on the site for something similar but was not happy with the results. Hence, I ask the question. I hope it is no dublicate.
 A: Wasn't me suggestion is perfectly fine, but you may also exploit
$$\frac{1}{(n^2+m^2)^{3/2}}=\int_{0}^{+\infty}\frac{x}{m}J_1(mx)e^{-nx}\,dx $$
where $J_1$ is a Bessel function of the first kind. This leads to
$$ \sum_{n\geq 1}\frac{1}{(n^2+m^2)^{3/2}} = \frac{1}{m}\int_{0}^{+\infty}\frac{x}{e^x-1}J_1(mx)\,dx $$
and it is enough to show that $\int_{0}^{+\infty}\frac{x}{e^x-1}J_1(mx)\,dx$ is $O(m^{-\varepsilon})$ for some $\varepsilon > 0$.
By invoking the full power of Tricomi's approximation
$$ J_1(x) \approx \frac{\sin(x)-\cos(x)}{\sqrt{\pi x}} $$
(holding for any sufficiently large $|x|$) one gets $I(m)=\int_{0}^{+\infty}\frac{x}{e^x-1} J_1(mx)\,dx = O(m^{-1})$. It is also possible to derive the weaker $I(m)=O(m^{\varepsilon-1/2})$ by elementary tools, i.e. Cauchy-Schwarz and
$$ \int_{0}^{+\infty} e^{-x} J_1(mx)^2\,dx = \frac{(2m^2+1)K(-4m^2)-E(-4m^2)}{\pi m^2}\sim \frac{\log m}{\pi m}.$$

An even simpler way is to notice that, by the convexity of
$\frac{1}{(m^2+n^2)^{3/2}}$ and the Hermite-Hadamard inequality
$$ \sum_{m\geq 1}\frac{1}{(m^2+n^2)^{3/2}}\leq
 \int_{1/2}^{+\infty}\frac{dx}{(x^2+n^2)^{3/2}} =
 \frac{4}{\sqrt{4n^2+1}\left(1+\sqrt{4n^2+1}\right)}\leq
 \frac{1}{n^2}$$ holds for any $n\in\mathbb{N}^+$.

A: Since you want an elementary proof, let's look for one even simpler, without any estimates for number-theoretic functions:
$2\,(m^2+n^2)=(m+n)^2+(m-n)^2\ge(m+n)^2$. Thus,
$$\sum^\infty_{m,n\ge1}\frac1{(m^2+n^2)^{3/2}}\le2^{3/2}\sum^\infty_{m,n\ge1}\frac1{(m+n)^3}.$$
Now $$\sum^\infty_{m,n\ge1}\frac1{(m+n)^3}=\sum_{k\ge2}\sum_{m+n=k}\frac1{k^3}=\sum_{k\ge2}\frac{k-1}{k^3}=\zeta(2)-\zeta(3),$$
so $$\sum^\infty_{m,n\ge1}\frac1{(m^2+n^2)^{3/2}}\le2^{3/2}\,(\zeta(2)-\zeta(3))<\infty.$$
REMARK: The numerical value of this estimate is $1.25\ldots$. Given that the exact value of the infinite sum is
$$\sum^\infty_{m,n\ge1}\frac1{(m^2+n^2)^{3/2}}=\zeta(3/2)\,\beta(3/2)-\zeta(3)=1.056348517615643291032890658317814644110805053803100504331\ldots,$$
that's not half bad. Here, $\beta(x)=\operatorname{L}_{-4}(x)=\zeta(x,1/4)-\zeta(x,3/4)$ is not exactly an elementary function.
A: Following the suggestion of wasn't me, we proceed by summation by parts. We have
$$
\sum_{N\geq 2}\frac{r_2(N)}{N^{3/2}}=\sum_{k\geq 2}\left(\frac{1}{k^{3/2}}-\frac{1}{(k+1)^{3/2}}\right)\sum_{N\geq 2}^kr_2(N)\,.
$$
As noted by wasn't me, the sum $\sum_{N\geq 2}^k r_2(N)$ is bounded by $\frac{\pi}{4}k<k$. The difference in each summand is
$$
=\frac{(1+\frac{1}{k})^{3/2}-1}{(k+1)^{3/2}}<\sum_{n=1}^\infty{3/2\choose n}k^{-n-3/2}
$$
In total, we get that the desired sum is bounded by
$$
\sum_{n=1}^\infty{3/2\choose n}(\zeta(n+1/2)-1)
$$
and this sum converges by the Leibniz criterion. Indeed, $\zeta(n+1/2)-1$ is positive and bounded by $\zeta(3/2)$ for all $n\geq 1$ and ${3/2\choose n}$ is an alternating monotonic null sequence for $n\geq 2$.
