double sequence of real numbers Which of the following statements are true?


*

*Let $\{a_{mn}\}, m ,n \in \mathbb{N}; $be an arbitrary double sequence of real numbers.
Then  $$\sum^ \infty_{m=1} \sum^ \infty_{n=1} a^3_{mn} = \sum^ \infty_{n=1} \sum^ \infty_{m=1} a^3_{mn} $$

*Let $\{a_{mn}\}, m ,n \in\mathbb{N}; $be an arbitrary double sequence of real numbers.
Then  $$\sum^ \infty_{m=1} \sum^ \infty_{n=1} a^2_{mn} = \sum^ \infty_{n=1} \sum^ \infty_{m=1} a^2_{mn} $$

*Let $\{a_{mn}\}, m ,n \in \mathbb{N}; $be an arbitrary double sequence of real numbers such that $|a_{mn} |\leq \sqrt{\frac{m}{n}}    $
Then  $$\sum^ \infty_{m=1} \sum^ \infty_{n=1} \frac{a_{mn}}{m^2n} = \sum^ \infty_{n=1} \sum^ \infty_{m=1} \frac{a_{mn}}{m^2n} $$
 A: Some HINTS:


*

*Let $a_{nn}=n$ and $a_{n+1,n}=-n$ for $n\in\Bbb Z^+$, and let $a_{mn}=0$ in all other cases.

*This is true, assuming that you allow $\infty$ as a sum; how you prove it will depend on what you already know.

*$\dfrac{\sqrt{m/n}}{m^2n}=\dfrac1{\sqrt{m^3n^3}}=\dfrac1{m^{3/2}n^{3/2}}$, so for fixed $m$ the series $\displaystyle\sum_{n\ge 1}\frac{a_{mn}}{m^2n}$ is absolutely convergent, and for fixed $n$ the series $\displaystyle\sum_{m\ge 1}\frac{a_{mn}}{m^2n}$ is absolutely convergent.
A: Use the Fubini–Tonelli theorem.  We have that $a_{mn}\leq\sqrt{\frac{m}{n}}= m^{\frac{1}{2}}\cdot n^{-\frac{1}{2}}$ implies $\frac{a_{mn}}{m^{\frac{1}{2}}\cdot n^{-\frac{1}{2}}}$ and
$$
\frac{a_{mn}}{m^{2}\cdot n^{1}}
= 
\frac{a_{mn}}{ m^{\frac{1}{2}}\cdot n^{-\frac{1}{2}}}
\cdot 
\frac{ m^{\frac{1}{2}}\cdot n^{-\frac{1}{2}}}{m^2n^1}
\leq 
\dfrac{1}{m^{\dfrac{3}{2}}}\cdot\dfrac{1}{n^{\dfrac{3}{2}}}
$$
Note that  $\displaystyle\sum_{n=1}^{\infty}\frac{1}{  n^{3/2}   }< \infty $ and $\displaystyle\sum_{m=1}^{\infty}\frac{1}{  m^{3/2}   }< \infty $ implies  $\displaystyle\sum_{n=1}^{\infty}\frac{1}{m^{3/2}  n^{3/2}   }< \infty $ and $\displaystyle\sum_{m=1}^{\infty}\frac{1}{  m^{3/2}n^{3/2}   }< \infty $ for all $m,n\in\mathbb{N}^*$.
Set the measures in $\mathbb{N}$:
$$
\mu(\{m\})=\frac{1}{m^{2}} \mbox{ and } \nu(\{n \})=\frac{1}{n}.
$$
The above observations it follows that the hypotheses of the theorem Fubini-Tonelli theorem are satisfied. Therefore we have the following.
\begin{align}
\sum^ \infty_{m=1} \sum^ \infty_{n=1} \frac{a_{mn}}{m^2n}=
&
\int_{\mathbb{N}}\int_{\mathbb{N}} a(m,n) d\mu(m) d\nu(n)
\\
=
&
\int_{\mathbb{N}}\int_{\mathbb{N}} a(m,n)  d\nu(n) d\mu(m)
\\
=
&
\sum^ \infty_{n=1} \sum^ \infty_{m=1} \frac{a_{mn}}{m^2n} 
\end{align}
