Show scalar multiplication sequence of a Cauchy sequence is also a Cauchy sequence. Let $(x_n)$ be a Cauchy sequence in a normed vector space $V$ and let $(\lambda_n)$ be a convergent sequence in $\mathbb{R}$. Show sequence $(\lambda_n x_n)$ is also Cauchy sequence. 
My try:
Since $(x_n)$ is a Cauchy sequence, for any given $\epsilon$, there is $N \in \mathbb{N}$ such that for all $n,m \geq N$ we have 
$$
\|x_n - x_m\| <  \epsilon
$$
We need to show
$$
\|\lambda_n x_n - \lambda_m x_m\| <  \epsilon
$$
 A: Consider the following identity: $$x_ny_n - x_my_m = (x_n-x_m)(y_n - y_m) + x_m(y_n - y_m) + y_m(x_n - x_m) $$ Everything should follow from this pretty quickly.
A: Use the usual trick. Consider
\begin{align*}
\|\lambda_n x_n - \lambda_m x_m\| &= \|\lambda_n x_n - \lambda_n x_m + \lambda_n x_m - \lambda_m x_m\| \\
&\le |\lambda_n| \|x_n - x_m\| + |\lambda_n - \lambda_m|\|x_m\|.
\end{align*}
Now, recall that Cauchy sequences are bounded. There must exist $A, B$ such that $\|x_n\| \le A$ and $|\lambda_n| \le B$ for all $n$. Thus,
$$\|\lambda_n x_n - \lambda_m x_m\| \le A\|x_n - x_m\| + B|\lambda_n - \lambda_m|.$$
Both terms in this sum can be made less than $\varepsilon / 2$ easily using the Cauchiness of the two sequences $(x_n)$ and $(\lambda_n)$.
A: Let $(V,\|.\|)$ be normed vector space, and $(x_n)_{n \in \mathbb N} \in V^\mathbb N$ be a cauchy sequence. Moreover, let $(\lambda_n)_{n \in \mathbb N} \in \mathbb R^\mathbb N$ be a convergent sequence with limit $\lambda \in \mathbb R$.
First claim: Both cauchy and convergent sequences are bounded.
For convergent, we have that for some $\epsilon > 0$ there exists $N(\epsilon)$ such that $|\lambda_n - \lambda | < \epsilon$ for $n > N(\epsilon)$.
Clearly then for every $n \in \mathbb N$ we have $|\lambda_n| \le \max\{|\lambda_1|,...,|\lambda_n|, |\lambda| + \epsilon \}$.
For cauchy one, if it wasn't bounded, then we can choose a subsequence $(n_k)_{ k \in \mathbb N}$ such that $|x_{n_k}| > k$ for every $k \in \mathbb N$. Now taking any $\epsilon > 0$, we can't find such $N(\epsilon)$, such that $|x_{n_k} - x_{n_r}| < \epsilon$, for $k,r \ge N(\epsilon)$ because when we take any $N(\epsilon)$, then there exist such $r \in \mathbb N$ so that $|x_{n_r}| > |x_{n_{N(\epsilon)}} | + \epsilon + 1$.
Now as we have this,  we can proceed as follows:
Take any $\epsilon > 0$, and let $B$ be a bound for cauchy sequence. We want to show, there exists $N(\epsilon)$, such that for $n,m > N(\epsilon)$ we have $\| \lambda_n x_n - \lambda_m x_m \| < \epsilon$.
Firsly take such $M_1(\epsilon)$, so that for $n > M_1(\epsilon)$ we have $\lambda_n = \lambda + r_n$, where $|r_n| < \frac{\epsilon}{3B}$.
Secondly, take such $M_2(\epsilon)$ so that $\| x_n - x_m \| < \frac{\epsilon}{3\lambda}$ for $n,m > M_2(\epsilon)$
Take $N(\epsilon) = \max\{M_1(\epsilon),M_2(\epsilon)\}$
Then we have:
$ \|x_n\lambda_n - x_m\lambda_m\| = \| x_n(\lambda + r_n) - x_m(\lambda + r_m) \| \le \lambda \|x_n - x_m\| + \|x_nr_n\|+ \|r_mx_m\| \le \lambda \frac{\epsilon}{3\lambda} + B\frac{\epsilon}{3B} + B \frac{\epsilon}{3B} =\epsilon$
