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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 $$\sum^ \infty_{m=1} \sum^ \infty_{n=1} \frac{a_{mn}}{m^2n} = \sum^ \infty_{n=1} \sum^ \infty_{m=1} \frac{a_{mn}}{m^2n} $$

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the last one seems to converge absolutely –  Santosh Linkha Jan 10 '13 at 17:40
    
If the terms are positive (your second question), see here: math.stackexchange.com/questions/89814/… –  Martin Sleziak Jan 23 at 13:58

2 Answers 2

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.

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still not getting... –  Prasanta Jan 10 '13 at 17:43
    
@Prasanta: Which part? –  Brian M. Scott Jan 10 '13 at 17:45
    
both 1 and 2... –  Prasanta Jan 10 '13 at 17:47
    
@Prasanta: For (1) just evaluate the two sums for the double series that I suggested. For (2) and (3) you need to know that equality holds if the double series is absolutely convergent. –  Brian M. Scott Jan 10 '13 at 17:50
    
Maybe you intended a different sequence for 1. With your choice both sums equal $0$. Am I right? –  P.. Jan 10 '13 at 17:54

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 nostemos 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}

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