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Is there an infinite matrix $A_{mn}$ such that $\lim\limits_{n \to \infty }A_{mn}=0 $ for every $m$ and $\lim\limits_{m \to \infty }A_{mn}=1 $ for every $n$ ?

Any clue as to how to start on this?

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P.S: my latex tends to begin in a new line after the text whenever i inser $$ , any idea as to how to avoid that? –  Bhargav Aug 14 '11 at 2:24
    
Just use a single $ instead. –  Calle Aug 14 '11 at 2:29
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@Bhargav: That's what LaTeX always does. For in-line formulas, use a single $. –  Arturo Magidin Aug 14 '11 at 2:30
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@Bhargav: This isn't so much about an infinite matrix as it is about an infinite series with double indices. The fact that it's a matrix is irrelevant. Just take $A_{mn} = \frac{m}{m+n}$. –  Arturo Magidin Aug 14 '11 at 2:31
    
Ya now i get it , rather than looking at it fromt he point of a matrix question ,i should have seen t as a calculus quetsion then the problem would have been solved. Thx arturo and zev –  Bhargav Aug 14 '11 at 2:37

2 Answers 2

up vote 6 down vote accepted

How about $A_{mn}=\left(\frac{m}{m+1}\right)^n$? We get

$$\lim_{n\to\infty}A_{mn}=\lim_{n\to\infty}\left(\frac{m}{m+1}\right)^n=0$$ because $0<\frac{m}{m+1}<1$ for all $m\geq 1$, and $$\lim_{m\to\infty}A_{mn}=\lim_{m\to\infty}\left(\frac{m}{m+1}\right)^n=\left(\lim_{m\to\infty}\frac{m}{m+1}\right)^n=1^n=1.$$

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Another example:

  • $A_{mn}=0$ if $m<n$,
  • $A_{mn}=1$ if $m\geq n$.
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wow, you don't even have to know much about limits for this! –  GEdgar Aug 14 '11 at 13:05
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Exercise for the OP: you can even modify Jonas's answer to get a yet different limit for all "diagonals", that is, $\lim_{m \to \infty} A_{m,m+k} = \lim_{n\to\infty} A_{n+k,n} = \frac12$ for $k\geq 0$ –  Willie Wong Aug 14 '11 at 13:51
    
@Willie: Nice exercise! (I had to fight my urge to post a solution.) –  Jonas Meyer Aug 14 '11 at 18:12

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