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Suppose {$a_n$}$_{n=1}^{\infty}$ {$b_n$}$_{n=1}^{\infty}$ are sequences such that {$a_n$}$_{n=1}^{\infty}$ and {$a_n + b_n$}$_{n=1}^{\infty}$ converge. Prove that {$b_n$}$_{n=1}^{\infty}$ also converges.

I'm confused on a few parts. Taking what can be assumed, we know that for every $\epsilon$>0, there exists $N_1 \in \Bbb N$ such that for every n $\in \Bbb N$, if n$\ge$$N_1$, then |$a_n -A|\lt \epsilon$. Also, we know that for every $\epsilon$>0, there exists $N_2 \in \Bbb N$ such that for every n $\in \Bbb N$, if n$\ge$$N_2$, then |$a_n + b_n - (A+B)|\lt \epsilon$.

How would you put the two of these together in order to form a proof that {$b_n$}$_{(n=1)}^\infty$ converges too?

Please don't write a proof for the answer. Just give arguments that will help lead me on the path. Thanks!

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Hint: If $a_n\to a$ then $-a_n\to-a$. Show that $(a_n+b_n)+(-a_n)$ is the sum of two convergent sequences and so it converges as well.

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  • $\begingroup$ Is this the idea you're saying? {$a_n$}$_{n=1}^\infty$ converges to A; {$a_n + b_n$}$_{n=1}^\infty$ converges to . Thus, ($a_n + b_n$) + ($-a_n$) $\to$ C - A. And from here use the definition of convergence to show then {$b_n$} must also converge? $\endgroup$ – Student Feb 19 '13 at 2:09
  • $\begingroup$ Pretty much, yeah. $\endgroup$ – Asaf Karagila Feb 19 '13 at 5:38
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Use the fact that if $a_n\to a$ and $b_n \to b$, $a_n - b_n \to a - b$.

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  • $\begingroup$ how can you use $b_n$ converges to b if it isn't something that can be assumed from the hypothesis. $\endgroup$ – Student Feb 19 '13 at 1:46
  • $\begingroup$ Apply this to the sequences $a$ and $a + b$. $\endgroup$ – ncmathsadist Feb 19 '13 at 1:50

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