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

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.

• Show, generally, that $a_n \to a$ and $b_n \to b$ implies $a_n+b_n \to a+b$. Then, you get $x_n \to \frac{L+M}{2}$ (+, not -), and $y_n \to \frac{L-M}{2}$ (since $a_n-b_n = a_n+(-b_n) \to a+(-b) = a-b$). Commented Sep 22, 2019 at 19:57
• Just interchange the letters x and y, and interchange the letters L and M, in all of it. Commented Sep 22, 2019 at 20:01
• what's wrong with what you did is you said $\frac{-L-M}{2} = -\frac{L-M}{2}$. Be careful putting minus signs in numerators. Commented Sep 22, 2019 at 20:24

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.

• That makes sense. I was making it too complicated. Instead I could just go $(x_n + y_n) + (x_n - y_n) \rightarrow L+ M \Rightarrow 2 x_n \rightarrow L+M \Rightarrow x_n \rightarrow \frac{L + M}{2}$, and similarly for $y_n$ using the difference. I must admit though, I remain curious about what was wrong with my original work, since apparently I got the wrong answer for what the limit is. Commented Sep 22, 2019 at 20:08

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 $$y_n$$

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 $$j$$.