If both the sum and the difference of two sequences converge, do the individual sequences converge?

Question

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.

Answer 1

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.

Answer 2

Since $x_n+y_n \to l$ and $x_n-y_n \to s$

Then $2x_n=(x_n+y_n)+(x_n-y_n) \to l+s \Longrightarrow x_n \to \frac{l+s}{2}$

Similarly proove the convergence of $y_n$

Answer 3

If $\{x_n + y_n\}$ and $\{x_n - y_n\}$ converge.

Then $\{(x_n +y_n) + (x_n - y_n)\}$ and $\{(x_n + y_n) - (x_n + y_n)\}$ converge.

And so do $\{k[(x_n +y_n) + (x_n - y_n)]\}$ and $\{j[(x_n +y_n) - (x_n - y_n)]\}$ for any constants $k$ and $j$.