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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?2011-08-14
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    Just use a single $ instead.2011-08-14
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    @Bhargav: That's what LaTeX *always* does. For in-line formulas, use a single `$`.2011-08-14
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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 $A_{mn} = \frac{m}{m+n}$.2011-08-14
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    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 zev2011-08-14

2 Answers 2

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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
  • $A_{mn}=1$ if $m\geq n$.
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    wow, you don't even have to know much about limits for this!2011-08-14
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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$2011-08-14
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    @Willie: Nice exercise! (I had to fight my urge to post a solution.)2011-08-14