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I want to show for $a_n\rightarrow a$ and $b_n$ convergent, if $a_n+b_n\rightarrow a$, then $b_n\rightarrow 0$.

$\textbf{Pf.}$ Suppose $a_n+b_n\rightarrow a$. Then, for all $\epsilon>0$, $\exists$ $N$ with $|a_n+b_n-a|<\epsilon$. When I use the triangle inequality, I can brea up the abs value and I end up with $|b_n|$ by itself and I don't know how to complete it.

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$b_n = (a_n+b_n) - a_n$ and so $\lim b_n = \lim (a_n+b_n) - \lim a_n= a-a=0$.

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Hint: prove that if $a_n\to a$ and $b_n\to b$ then $a_n+b_n\to a+b$. Then note that limits are unique (or prove this) and then $a+b=a$ and the result follows

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  • $\begingroup$ Ok, that works. Is there a direct approach to the proof? $\endgroup$ – Moz Oct 23 '15 at 1:11
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    $\begingroup$ @Moz, yes just take Ihf's approach. For any $\varepsilon > 0$, there exists $N_1 \in \mathbb N$ such that for all $n \geq N$, $|a_n + b_n - a| < \varepsilon$ and $N_2 > N_1$ such that $n > N_2$ implies $|a_n - a| < \varepsilon$. By the triangle inequality, $n > N_2$ implies $|b_n| = |a_n + b_n - a + a - a_n| \leq |a_n + b_n - a| + |a - a_n| < \frac{\varepsilon}{2} \cdot 2 = \varepsilon.$ $\endgroup$ – user217285 Oct 23 '15 at 1:33
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If you want to do this by definition, let $\varepsilon >0$ and $\varepsilon' = \frac{\varepsilon}{2}$. We know there exists integers $N,M$ such that $$\left|a_n-a\right| = \left|a-a_n\right| <\varepsilon' \quad \left|(a_m+b_m)-a\right|<\varepsilon'$$ Take $N \geq M$ without loss of generality so that $\left|(a_n+b_n)-a\right|<\varepsilon'$. Then $$\begin{align}\varepsilon &= \varepsilon'+\varepsilon' \\ &> \left|a-a_n\right|+ \left|(a_n+b_n)-a\right| \\ &\geq \left|(a-a_n)+[(a_n+b_n)-a]\right| \tag{Triangle Inequality} \\ &= \left|b_n+(a-a)+(a_n-a_n)\right| \\ &= \left|b_n-0\right|\end{align}$$

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