Equivalence of convergence in metric spaces 
Let $\{x^{(k)}\}$ be a sequence in $\mathbb{R}^n$ with $x^{(k)} = (x_1^{(k)}, x_2^{(k)}, \ldots, x_n^{(k)})$, and $x = (x_1, x_2, \ldots, x_n) \in \mathbb{R}^n.$ 
  Then $\{x^{(k)}\}$ converges to $x$ with respect to the $l^2$-metric $\rho_2$ if and only if $\{x^{(k)}\}$ converges to $x$ with respect to the $l^1$-metric $\rho_1$.

Well, I need to show that for all $\epsilon > 0$, there exists an $N$ such that $||x^{k}-x||_2 < \epsilon$ if and only if $||x^{k}-x||_1 < \epsilon$. 
We have that $\rho_2(x^{(k)},x) = ||x^{k}-x||_2 = (\sum |x_j^{(k)} - x_j|^2)^{1/2}$, and that   $\rho_1(x^{(k)}, x) = ||x^{k}-x||_1 = \sum |x_j^{(k)} - x_j|$.
Previously, we showed that convergence of sequences for metric spaces is equivalent if we can find a constant $c > 1$ such that $\frac{1}{c} \rho_p(x,y) \leq \rho_q(x,y) \leq c\rho_p(x,y)$, so I thought maybe finding a $c$ such that $\frac{1}{c} \rho_2(x,y) \leq \rho_1(x,y) \leq c\rho_2(x,y)$ would be easier than playing with epsilons, but I can't find such a c. Can I have some help?
 A: Your intuition is right.  Let's see how to find these constants.  We'll use the Cauchy-Schwarz inequality on $\mathbb{R}^n$, which tells us that $| a \cdot b | \le \Vert a \Vert_2 \Vert b \Vert_2$ for all $a,b \in \mathbb{R}^n$.
To start off we use Cauchy-Schwarz to estimate
$$
\Vert x \Vert_1 = \sum_{i=1}^n |x_i|  = \sum_{i=1}^n |x_i | 1 \le \left(\sum_{i=1}^n |x_i|^2 \right)^{1/2} \left(\sum_{i=1}^n 1^2 \right)^{1/2} = \sqrt{n} \Vert x \Vert_2
$$
for any $x \in \mathbb{R}^n$.  
Next we note that 
$$
|x_j| \le \sum_{i=1}^n |x_i| = \Vert x \Vert_1 \text{ for }j=1,\dotsc,n
$$
and hence
$$
\max_{1\le j \le n} |x_j| \le \Vert x \Vert_1.
$$
In turn we use this to estimate
$$
\Vert x \Vert_2^2 = \sum_{i=1}^n |x_i|^2 \le \left( \sum_{i=1}^n |x_i| \right) \left(\max_{1\le j \le n} |x_j|  \right) \le \left( \sum_{i=1}^n |x_i| \right)^2,
$$
which tells us (after taking square roots) that 
$$
\Vert x\Vert_2 \le \Vert x \Vert_1.
$$
Combining the above we now know that 
$$
\Vert x \Vert_2 \le \Vert x \Vert_1 \le \sqrt{n} \Vert x \Vert_2
$$
for all $x \in \mathbb{R}^n$.  Using this, you can complete the sketch of your argument.  
