# Will a convergent Double Sequence be bounded also?

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 .

• 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 at 16:28
• What is the definition of $\lim_{m,n\to \infty }a_{mn}$ ? – zhw. Apr 26 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 at 1:49
• Then Can you please tell me will the sequence $1 /m + 1/n$ be convergent or not?@CalvinKhor – cmi Apr 27 at 1:50