# Show that $(3x_{n}+4y_{n})$ is also Cauchy sequence.

Show that if $(x_{n})$ and $(y_{n})$ are Cauchy sequences in $X$, then the sequence $(3x_{n}+4y_{n})$ is also Cauchy sequence using the definition of a Cauchy sequence.

Attempt

Let $\epsilon > 0$ be given.

By definition of a Cauchy sequence
$\forall\epsilon>0:\exists N_{1}\in\mathbb{N}:n,m> N_1\implies|3x_{n}-3x_{m}|<\frac{\epsilon}{2}$
$\forall\epsilon>0:\exists N_{2}\in\mathbb{N}:n,m> N_2\implies|4y_{n}-4y_{m}|<\frac{\epsilon}{2}$

Let $N_{\epsilon}$ = max{$N_{1}$,$N_{2}$}

Then $\forall\epsilon>0:\exists N_{\epsilon}\in\mathbb{N}:n,m> N_{\epsilon}\implies|(3x_{n}+4y_{n}) - (3x_{m}+4y_{m})| =|(3x_{n}-3x_{m}) + (4y_{n}-4y_{m})|< |3x_{n}-3x_{m}| + |4y_{n}-4y_{m}|<\frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Thus $(3x_{n}+4y_{n})$ is also Cauchy sequence.

• That is correct, although I would write $|x_n-x_m|<\frac\varepsilon6$, $|y_n-y_m|<\frac\varepsilon8$ to make it more clear. – Math1000 Oct 18 '15 at 15:27