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This is the same question as this. However, my question is how to pursue the proof that the series diverges using the Theorem

Suppose $a_{m,n}\in[0,\infty]$ for each $(m,n)\in\mathbb{N}\times\mathbb{N}$ and that $f$ is any one to one mapping of $\mathbb{N}$ onto $\mathbb{N}\times\mathbb{N}$. Then $$ \sum_{m=1}^\infty\sum_{n=1}^\infty a_{m,n} = \sum_{k=1}^\infty a_{f(k)}$$

and $f^{-1}$ as

$$ f^{-1}(m,n) = (m+n-1)(m+n)/2-(m-1). $$

The series is

$$ \sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{(m+n)^2}=\infty. $$

I tried to find an inequality that helps to prove this. For example $1/(m+n)^2>1/((m+n)^2+(m+n))$, but with this I cannot use $f$.

user2820579
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  • $f$ is supposed to be a map from $\mathbb N$ to $\mathbb N\times \mathbb N$. What you have defined is $f^{-1}$. – Kavi Rama Murthy Jul 01 '20 at 23:14
  • Fixed now. Fixed. – user2820579 Jul 01 '20 at 23:19
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    Use a one to one function that maps the first $n^2$ numbers into $(i, j), 1\leq i, j \leq n$ then the sum of the first $n^2$ terms is $\sum_{i=1}^n i/(i+1)^2+i/(2n-i+1)^2$ because to get $i+1$ as a sum of $m+n$ where $1 \leq m, n\leq n$ you can do that in $i$ ways, similarly for $2n-i+1$ now the sum diverges by comparison with the harmonic sum. – kingW3 Jul 01 '20 at 23:48

1 Answers1

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Too long for a comment.

Since in comments, you received a good answer from @kingW3, I wondered about the asymptotics of $$S_p=\sum_{m=1}^p\sum_{n=1}^p\frac{1}{(m+n)^2}$$ $$\sum_{n=1}^p\frac{1}{(m+n)^2}=\psi ^{(1)}(m+1)-\psi ^{(1)}(m+p+1)$$ $$S_p=2 \psi ^{(0)}(p+2)-\psi ^{(0)}(2 p+2)+2 (p+1) \psi ^{(1)}(p+2)-(2 p+1) \psi ^{(1)}(2 p+2)-\frac{\pi ^2}{6}+\gamma$$ Now, using asymptotics $$S_p=\left(\gamma +1-\frac{\pi ^2}{6}-\log (2)\right)+\log(p)+\frac{3}{2 p}-\frac{35}{48 p^2}+O\left(\frac{1}{p^3}\right)$$ while $$H_p=\gamma +\log (p)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$

The expansion for $S_p$ seems to be a quite good approximation even for small values of $p$.

For example $S_{10} \approx 1.68473$ while the above truncated expansion gives $1.68443$.

Claude Leibovici
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