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?
 A: 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*}
A: 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.
A: 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 $n,m > \max(N,M).$ 
