# If $|a_{n_j} - b_j| < 1/j$ and $b_j \rightarrow b$ then $a_{n_j}\rightarrow b$

I have some sequence $(a_n)$ with a subsequence $(a_{n_j})$ and another sequence $(b_j)$. How do I formally justify the fact that if $|a_{n_j} - b_j| < 1/j$ and $b_j \rightarrow b$ as $j$ approaches infinity then $a_{n_j}$'s limit as $j$ approaches infinity is also $b$?

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It seems like the sequence $A_n$ doesn't matter, only the subsequence $\{a_{n_j} : j \in \mathbb{N}\}$. Am I reading the question correctly? –  Trevor Wilson Feb 1 '13 at 0:50

$0\leq |a_{n_j}-b_j|<\frac{1}{j}\to 0$, so $|a_{n_j}-b_j|\to 0$ by squeezing.
$0\leq |a_{n_j}-b|\leq |a_{n_j}-b_j|+|b_j-b|\to 0+0=0$, so $|a_{n_j}-b|\to 0$ by squeezing again.
This is simply an application of the triangle inquality. Let $\epsilon>0$. Let $J$ be large enough such that $j\geq J$ implies that $1/j<\epsilon/2$ and such that $j\geq J$ implies that $|b_j-b|<\epsilon/2$. Then we let $j\geq J$. We have that $|a_{n_j}-b|=|(a_{n_j}-b_j)-(b_j-b)|\leq |a_{n_j}-b_j|+|b_j-b|<\epsilon/2+\epsilon/2=\epsilon$. This implies that $a_{n_j}\rightarrow b$.