# Show that the sequence $\langle b_n\rangle$ Converges to $1$

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 := $$\lim_{m\to\infty} a_{mn} exists. 2. \forall increasing sequences \langle m_k\rangle and \langle n_k\rangle of positive integers \lim_{k\to\infty} a_{m_k n_k} =1 Show that sequence \langle b_n\rangle converges to 1. • I changed \lim_{k\to\infty} a_{m_k n_k} =1 to \lim_{k\to\infty} a_{m_k n_k} =1 and <b_n> to \langle b_n\rangle, and <b_n> to \langle b_n\rangle. That is standard usage. {}\qquad{} – Michael Hardy Jun 22 '15 at 13:37 • @MichaelHardy Thank you.Regards, – Dontknowanything Jun 22 '15 at 13:39 ## 2 Answers Given any \epsilon > 0, by condition 1, we can choose an increasing sequence \{m_k\} such that$$\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.$$• Thank you for your response. – Dontknowanything Jun 24 '15 at 4:19 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).

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