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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?

YuiTo Cheng
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cmi
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    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 $0$ – Calvin Khor Apr 26 '19 at 16:28
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    What is the definition of $\lim_{m,n\to \infty }a_{mn}$ ? – zhw. Apr 26 '19 at 17:38
  • for every $e > 0$ there exists a natural number $N$ such that $|a_{mn} - l| < e$ for all $m ,n > N$..@zhw. – cmi Apr 27 '19 at 01:49
  • Then Can you please tell me will the sequence $1 /m + 1/n$ be convergent or not?@CalvinKhor – cmi Apr 27 '19 at 01:50
  • I took help from this site http://mathonline.wikidot.com/double-sequences-of-real-numbers-review...@zhw. – cmi Apr 27 '19 at 06:01
  • Can you please clarify whether the statement is correct or wrong? I am still waiting for your reply..@zhw. – cmi May 01 '19 at 18:00
  • Please see https://math.stackexchange.com/questions/3204124/what-is-the-definition-of-double-sequence-a-mn-being-convergent-to-l?noredirect=1#comment6592661_3204124 also@CalvinKhor – cmi May 01 '19 at 18:05
  • it is convergent – rhombicosicodecahedron May 03 '19 at 18:01

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