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Prove that if real sequences $(x_n)$ and $(x_n + y_n)$ converge, then $(y_n)$ converges.

My attempt so far:

Suppose that the limits of $(x_n)$ and $(x_n + y_n)$ are $x$ and $x+y$ respectively. (Intuitively, $y_n \rightarrow y$ as $n \rightarrow \infty$.) Then given any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $|x_n - x| < 3\epsilon/2$ and $|(x_n + y_n) - (x+y)| < \epsilon/2$ for all $n > N$.

By triangle inequality, we have that $|(x_n + y_n) - (x+y)| \leq |x_n - x| + |y_n - y|$. I want to use this relation and $|x_n - x|$ to obtain $|y_n - y| < \epsilon$. But I haven't had any luck so far.

Any help would be appreciated!

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From what you have, you can deduce $\newcommand{\ep}{\epsilon}|y_n-y|<2\ep$. Observe that $$|y_n-y|=|x_n+y_n-(x+y)-(x_n-x)|\le |x_n+y_n-(x+y)|+|-(x_n-x)|<\frac32\ep+\frac\ep2.$$ By choosing $\ep/2$ instead of $3\ep/2$ you could have obtained a final $\ep$ rather than $2\ep$.

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Set $L$ to be the limit of $(x_{n}+y_{n})$ (or we take $y=L-x$ as OP noted), and we claim that $y=L-x$ is the limit of $(y_{n})$: For $\epsilon>0$, choose $N$ such that for all $n\geq N$, $|x_{n}-x|<\epsilon/2$ and $|x_{n}+y_{n}-L|<\epsilon/2$, then for such an $n$, we have \begin{align*} |y_{n}-(L-x)|&=|y_{n}+x_{n}-L-x_{n}+x|\\ &\leq|x_{n}+y_{n}-L|+|x_{n}-x|\\ &<\epsilon. \end{align*}

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  • $\begingroup$ Thanks for your solution! The reason why I set the limit of $(x_n + y_n)$ to be $x+y$ is as follow: If the limit of $(x_n)$ is $x$ and the limit of $(x_n + y_n)$ is $L$, then there exists $y \in \mathbb{R}$ such that $L = x + y$. $\endgroup$ – user376127 Nov 29 '17 at 5:30
  • $\begingroup$ Okay, then it is my misunderstanding, sorry. $\endgroup$ – user284331 Nov 29 '17 at 5:31
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$$y_n = (x_n + y_n ) - x_n$$

Now, take limits of both sides and use that you can split up the limit on the right. You end up with a sum of $2$ real numbers, hence $(y_n)$ converges

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