# Does an infinite matrix exist with each row converging to 0 and each column to 1?

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 $$, any idea as to how to avoid that? – Bhargav Aug 14 '11 at 2:24 Just use a single  instead. – Calle Aug 14 '11 at 2:29 @Bhargav: That's what LaTeX always does. For in-line formulas, use a single . – Arturo Magidin Aug 14 '11 at 2:30 @Bhargav: This isn't so much about an infinite matrix as it is about an infinite series with double indices. The fact that it's a matrix is irrelevant. Just take A_{mn} = \frac{m}{m+n}. – Arturo Magidin Aug 14 '11 at 2:31 Ya now i get it , rather than looking at it fromt he point of a matrix question ,i should have seen t as a calculus quetsion then the problem would have been solved. Thx arturo and zev – Bhargav Aug 14 '11 at 2:37 ## 2 Answers How about A_{mn}=\left(\frac{m}{m+1}\right)^n? We get$$\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
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