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The following question was asked in my Masters entrance examination but unfortunately I was unable to answer this. Please tell me how to approach this problem correctly.

Suppose $\langle a_{mn}\rangle$ is a double sequence such that

$1$. $ \forall n$, $b_n := $$\lim_{m\to\infty} a_{mn}$ exists.

$2$. $ \forall$ increasing sequences $\langle m_k\rangle$ and $\langle n_k\rangle$ of positive integers $\lim_{k\to\infty} a_{m_k n_k} =1$

Show that sequence $\langle b_n\rangle$ converges to $1$.

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    $\begingroup$ I changed $\lim_{k\to\infty} a_{m_k n_k}$ =1 to $\lim_{k\to\infty} a_{m_k n_k} =1$ and <$b_n$> to $\langle b_n\rangle$, and $<b_n>$ to $\langle b_n\rangle$. That is standard usage. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 22 '15 at 13:37
  • $\begingroup$ @MichaelHardy Thank you.Regards, $\endgroup$ – Dontknowanything Jun 22 '15 at 13:39
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Given any $\epsilon > 0$, by condition 1, we can choose an increasing sequence $\{m_k\}$ such that $$\left|a_{m_k k} - b_k\right| < \epsilon/2.$$

By condition 2, the sequence $\{a_{m_k k}\}$ converges to $1$, that is, $\exists K \in \mathbb{N}$, such that for all $k > K$, we have $$\left|a_{m_k k} - 1\right| < \epsilon/2.$$ Now the result follows because for all $k > K$, $$\left|b_k - 1\right| \leq \left|b_k - a_{m_k k}\right| + \left|a_{m_k k} - 1\right| < \epsilon/2 + \epsilon/2 = \epsilon.$$

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  • $\begingroup$ Thank you for your response. $\endgroup$ – Dontknowanything Jun 24 '15 at 4:19
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If the limit does not exist or is not equal to $1$, then for every $\varepsilon>0$, no matter how small, either there are infinitely many values of $n$ for which $b_n<1-\varepsilon$ or there are infinitely many for which $b_n>1+\varepsilon$. In the former case, let $n_1,n_2,n_3,\ldots$ be a sequence for which $b_{n_k}$ is always (i.e. for all $k$) less than $1-\varepsilon$ and let $m_k=k$. See if you can prove $\liminf\limits_{k\to\infty} a_{m_k,n_k}\le 1-\varepsilon$ (I don't think you get get "$<$" rather than "$\le$" at that point, but "$\le"$ is good enough).

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  • $\begingroup$ Everything is clear.Thank you for your response! $\endgroup$ – Dontknowanything Jun 24 '15 at 4:18

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