# Prove $\{aX_n +bY_n\}$ is a Cauchy Sequence.

Let $\{X_n\}$ and $\{Y_n\}$ be Cauchy sequences and let $a$ and $b$ be nonzero real numbers. Prove that $\{aX_n$ + b$Y_n\}$ is a Cauchy sequence.

So far I have Since $\{X_n\}$ is Cauchy then for every $\epsilon>0$ there exists an $N_1 \in \mathbb{N}$ s.t. when $m,n\geq N_1$ then $|X_n-X_M|< \frac{\epsilon}{ab}$(?).

Since $\{Y_n\}$ is Cauchy then for every $\epsilon>0$ there exists an $N_2 \in \mathbb{N}$ s.t. when $m,n\geq N_2$ then $|Y_n-Y_M|< \frac{\epsilon}{ab}$(?). Choose $N=\max\{N_1,N_2\}$, so for every $\epsilon>0$, when $n,m\geq N$ then $|a(X_n+Y_n) - b(X_m+Y_m)|=|aX_n+aY_n-bX_n-bY_m|$ but something is missing?

• Since you are new to math stack exchange, you would probably benefit from taking a look at this handy guide to formatting mathematics: math.stackexchange.com/help/notation – TravisJ Apr 30 '15 at 1:57
• It's a silly technical point, but, you want to be careful about dividing by $a$ and $b$, as, they could be 0. – James Apr 30 '15 at 2:02
• okay thank you all, your answers really helped :) – Sandra Eades Apr 30 '15 at 2:23
• @James, either by edit (or by oversight--I've done it frequently) there is the hypothesis that $a$ and $b$ are non-zero. But you are right on that you must be careful to not divide by $0$... in this case, there is no worry. – TravisJ Apr 30 '15 at 3:09
• @TravisJ it was most likely oversight! Well spotted. – James Apr 30 '15 at 13:22

Assuming your sequences are in $\mathbb{R}$ then you know that they converge (because they are Cauchy). Say $x_{n}\to x$ and $y_{n}\to y$. Then $ax_{n}+by_{n}\to ax+by$. Convergent sequences are Cauchy.

If you really want to bring out the $\epsilon$'s, then you could say that since $x_n\to x$ there is an $N_x$ so that if $n\geq N_x$ then $|x_n - x|<\frac{\epsilon}{2|a|}$ Similarly, there is an $N_y$ so that for $n\geq N_y$ you have that $|y_n-y|<\frac{\epsilon}{2|b|}$. Then, if $n\geq \max\{N_x, N_y\}$ it follows that

\begin{align*} |ax_{n}+by_{yn}-(ax+by)| &\leq |ax_{n}-ax| + |by_{n}-by| \\ &= |a||x_n - x|+|b||y_n-y| \\ &<|a|\frac{\epsilon}{2|a|}+|b|\frac{\epsilon}{2|b|} \\ &=\epsilon \end{align*}

do it in two phases.

$X_n$ Cauchy implies $aX_n$ is.

$X_n, Y_n$ Cauchy implies that $X_n+Y_n$ is .

For the first, given $\epsilon >0,$ there exist $N$ st $m,n>N$ implies $$|X_n-X_m| < \epsilon/a.$$ So $$|aX_n-aX_m| < \epsilon,$$ and we are done.

For the second, find $M,N$ so that $$|X_n-X_m| < \epsilon/2.$$ with $n,m > M,$ $$|Y_n-Y_n| < \epsilon/2$$ with $n,m> N.$

Then the condition holds via the triangle inequality for $n,m > \max(N,M).$

Let $M=\max\{|a|,|b|\}$. Since $(x_n)$ and $(y_n)$ are Cauchy sequences, for every $\varepsilon>0$ there exists $n_1,n_2\in\mathbb{N}$ such that if $n,m>n_1,n_2$, $$|x_n-x_m|<\dfrac{\varepsilon}{2M}\ \ \ \text{and}\ \ \ |y_n-y_m|<\dfrac{\varepsilon}{2M}$$ Let $N=\max\{n_1,n_2\}$, then if $n,m>N$, \begin{align*} |(ax_n+by_n)-(ax_m+by_m)|&=|a(x_n-x_m)+b(y_n-y_m)|\\ &\leqslant |a||x_n-x_m|+|b||y_n-y_m|\\ &\leqslant M(|x_n-x_m|+|y_n-y_m|) \\ &< M\left(\dfrac{\varepsilon}{2M}+\dfrac{\varepsilon}{2M}\right)=\varepsilon \end{align*} So $(ax_n+by_n)$ is Cauchy.