Question: Which of the following statements are true?

1. Let $\{a\_{mn}\}, m ,n \in \mathbb{N}; $be an arbitrary double sequence of real numbers.
Then $$\sum^ \infty\_{m=1} \sum^ \infty\_{n=1} a^3\_{mn} = \sum^ \infty\_{n=1} \sum^ \infty\_{m=1} a^3\_{mn} $$
2. Let $\{a\_{mn}\}, m ,n \in\mathbb{N}; $be an arbitrary double sequence of real numbers.
Then $$\sum^ \infty\_{m=1} \sum^ \infty\_{n=1} a^2\_{mn} = \sum^ \infty\_{n=1} \sum^ \infty\_{m=1} a^2\_{mn} $$
3. 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 $$\sum^ \infty\_{m=1} \sum^ \infty\_{n=1} \frac{a\_{mn}}{m^2n} = \sum^ \infty\_{n=1} \sum^ \infty\_{m=1} \frac{a\_{mn}}{m^2n} $$

Answer: 
Some HINTS:

1. Let $a\_{nn}=n$ and $a\_{n+1,n}=-n$ for $n\in\Bbb Z^+$, and let $a\_{mn}=0$ in all other cases.
2. This is true, assuming that you allow $\infty$ as a sum; how you prove it will depend on what you already know.
3. $\dfrac{\sqrt{m/n}}{m^2n}=\dfrac1{\sqrt{m^3n^3}}=\dfrac1{m^{3/2}n^{3/2}}$, 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.
Use the [Fubini–Tonelli theorem](http://en.wikipedia.org/wiki/Fubini%27s_theorem#The_Fubini.E2.80.93Tonelli_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}