# $\lim_{n\rightarrow \infty} (a_n+b_n)=0$ $\lim_{n\rightarrow \infty} (b_n-c_n)=0$ imply $\lim_{n\rightarrow \infty} (a_n+c_n)=0$?

Consider some sequence $\{a_n\}_{n\in \mathbb{N}}<0$, $\{b_n\}_{n\in \mathbb{N}}>0$, $\{c_n\}_{n\in \mathbb{N}}>0$ and assume $$\begin{cases} \lim_{n\rightarrow \infty} (a_n+b_n)=0\\ \lim_{n\rightarrow \infty} (b_n-c_n)=0 \end{cases}$$ Does this imply $$\lim_{n\rightarrow \infty} (a_n+c_n)=0 \text{ ?}$$

I am confused because: $$\begin{cases} (1) \hspace{1cm}\lim_{n\rightarrow \infty} (a_n+b_n)=0 \text{ is equivalent to write } a_n=-b_n+o(1)\\ (2) \hspace{1cm}\lim_{n\rightarrow \infty} (b_n-c_n)=0\text{ is equivalent to write } b_n=c_n+o(1)\\ \end{cases}$$ Put (2) in (1) and get $$a_n=-c_n+o(1)+o(1)=-c_n+o(1) \text{ which is equivalent to write } \lim_{n\rightarrow \infty} (a_n+c_n)=0\\$$ where $o(1)$ is a number converging to zero as $n\rightarrow \infty$

What is wrong in my arguments?

Yes. First, assume $\lim_{n\to\infty} (a_n + c_n) = x$ where $x \ne 0$. Now add the second equation to get $$\lim_{n\to\infty} (a_n + c_n) + \lim_{n\to\infty} (b_n - c_n) = x + 0$$ $$\lim_{n\to\infty} (a_n + b_n + c_n - c_n) = x$$ $$\lim_{n\to\infty} (a_n + b_n) = x$$ Since we already know from the equations given that $\lim_{n\to\infty} (a_n + b_n) = 0$, this leads to a contradiction proving that $x = 0$ and thus that $\lim_{n\to\infty} (a_n + c_n) = 0$.

• The contradiction implies either $x=0$ or $x$ does not exist. Why have you assumed $x=0$ is the true branch, doesn't that need to be proven that the limit indeed exists? Jul 7, 2023 at 15:15

No.

You could have $a_n=-n$, $b_n=c_n=n$.

• Thanks, could you briefly explain why my arguments above are wrong? (see added part)
– Star
May 11, 2018 at 16:11
• Am I missing something here? Don't these values satisfy all 3 equations May 11, 2018 at 17:52
• I do not believe this answer applies any longer. May 12, 2018 at 1:01
• This answer is wrong Jul 7, 2023 at 15:14