Prove or else give a counterexample: If $x_n + y_n$ converges and if $x_n – y_n$ converges, then $x_n$ converges and $y_n$ converges.
My work so far:
Suppose that $x_n + y_n$ converges to $L$ and $x_n – y_n$ converges to $M$.
We know that there exists an $N_1$ in the natural numbers such that $n > N_1$ implies that $|(x_n + y_n) – L| < \epsilon$.
We know that there exists an $N_2$ in the natural numbers such that $n > N_2$ implies that $|(x_n – y_n) – M| < \epsilon$.
Let $N = \max\{N_1,N_2\}$. Then, for all $n > N$, we have
\begin{align}
&& |(x_n + y_n) – L| + |(x_n – y_n) – M| && < 2 \epsilon \\
-2 \epsilon < && x_n + y_n – L + x_n – y_n – M && < 2 \epsilon \\
-2 \epsilon < && 2 x_n – L – M && < 2 \epsilon \\
– \epsilon < && x_n – \frac{L – M}{2} && < \epsilon \\
&& \Big|x_n – \frac{L – M}{2}\Big| && < \epsilon
\end{align}
Therefore, $x_n$ converges to $\frac{L – M}{2}$.
My problem is trying to make this work for $y_n$ as well. I feel like it should be relatively simple to adapt this to show that $y_n$ converges, but I've been playing around with multiplying by -1 and the triangle inequality, and I can't seem to get it to come out right.
Best Answer
This is almost trivial: the sum, which is $2x_n$, also converges (to $L+M$), so $x_n\to\frac{L+M}{2}.$ For $y_n$ consider the difference.