A convergent Double Sequence will be bounded also.

My Attempt: I think the statement is not true.

Counter Example : $a_{1n} = n$, $ a_{mn} = 1/m + 1/n$ for all $m \geq 2$

lim$_{m,n \to \infty} a_{mn} =0$ But $ a_{mn}$ is not bounded .

Have I gone wrong anywhere? Can anyone please help me?

  • 3
    $\begingroup$ Your counterexample doesn't work, $\lim_{m,n\to\infty} a_{mn} = 0$ can be defined as for every $\epsilon>0$, there exists $N$ such that if $m+n>N$, then $$|a_{mn}| < \epsilon $$ and this is violated when $m=1,n=n$. Its not the same thing as $\lim_{n\to\infty} \lim_{m\to\infty} a_{mn}$, for which your $a_{mn}$ does give a value of $0$ $\endgroup$ – Calvin Khor Apr 26 at 16:28
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    $\begingroup$ What is the definition of $\lim_{m,n\to \infty }a_{mn}$ ? $\endgroup$ – zhw. Apr 26 at 17:38
  • $\begingroup$ for every $e > 0$ there exists a natural number $N$ such that $|a_{mn} - l| < e$ for all $m ,n > N$..@zhw. $\endgroup$ – cmi Apr 27 at 1:49
  • $\begingroup$ Then Can you please tell me will the sequence $1 /m + 1/n$ be convergent or not?@CalvinKhor $\endgroup$ – cmi Apr 27 at 1:50
  • $\begingroup$ I took help from this site mathonline.wikidot.com/…. $\endgroup$ – cmi Apr 27 at 6:01

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