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I think True.

What to show: $\exists b \in \mathbb R, \forall \epsilon > 0, \exists N_1>0$, such that for all $n \in \mathbb N$, if $n > N_1$, then $|b_n - b| < \frac{\epsilon}{2}$

Given: $\exists a \in \mathbb R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in \mathbb N$, if $n > N_2$, then $|a_n - b|< \frac{\epsilon}{4}$

Let $L \in \mathbb R$

Given: $|(a_n + b_n) - L| < \epsilon$

Let $\epsilon > 0$ be arbitrary

Choose N = $max(N_1, N_2) > 0$

Suppose $n > N$, then

let $b_n = (a_n + b_n) - a_n$

$$|(a_n + b_n) - a_n - (a + b) + a| = |(a_n - a) + (b_n - b) + (-a_n + a)|$$

$$\leq 2|a_n - a| + |b_n - b| \text{ By triangle inequality}$$

$$< 2\frac{\epsilon}{4} + \frac{\epsilon}{2}$$

$$= \epsilon$$

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Just we'll write $b_n=(a_n+b_n)-a_n$

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  • $\begingroup$ Is my edit what you meant? $\endgroup$ – Tinler Mar 22 '17 at 6:56
  • $\begingroup$ @Tinler Yes, of course! $\endgroup$ – Michael Rozenberg Mar 22 '17 at 7:04

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