I saw this question is asked previously but still the answer is not convincing to me. The theorem states that a sequence in the metric space $(\mathbb{R}^n,d_{l^1})$ converges if and only if it converges in $(\mathbb{R}^n, d_{l^2})$.
Proof for the second implication : Suppose $\lim_{k \to \infty} \sqrt{|x_1^{(k)}-x_1|^2+|x_2^{(k)}-x_2|^2+\cdots +|x_n^{(k)}-x_n|^2}=0$, i.e, the sequence $(x_1^{(k)},x_2^{(k)},\cdots, x_n^{(k)})_{k \to \infty}\to(x_1,x_2,\cdots, x_n)$
By the CS-inequality, $n{\sum a_i^2}\geq (\sum a_i)^2 $ so that $$\sqrt{|x_1^{(k)}-x_1|^2+|x_2^{(k)}-x_2|^2+\cdots +|x_n^{(k)}-x_n|^2} \geq \frac{|x_1^{(k)}-x_1|+\cdots + |x_n^{(k)}-x_n|}{n}\geq 0$$ and by the Squeeze Theorem $\frac{1}{n}\lim_{k \to \infty} ||x_1^{(k)}-x_1|+\cdots +|x_n^{(k)}-x_n||=0$ which gives the desired conclusion.
I would be thankful if you check this proof and also suggest whether i should start with metric spaces or abstract algebra thoroughly. Thanks.
Take $x \in \mathbb{R}^n$, $x=(x_1,\ldots,x_n)$. As you said, by Cauchy-Scharz inequality for one side, and by looking a bit closer for the other side, one can show that \begin{align} \frac{1}{n}\|x\|_1 \leqslant \|x\|_2 \leqslant \sqrt{n}\|x\|_1 \end{align} Say $(x_k)$ is a sequence and $x_{\infty}$ a point. Then \begin{align} \frac{1}{n}\|x_k-x_{\infty}\|_1 \leqslant \|x_k-x_{\infty}\|_2 \leqslant \sqrt{n}\|x_k-x_{\infty}\|_1 \end{align} You can conclude from that the following \begin{align} \|x_k-x_{\infty}\|_1 \to_{k\to\infty} 0 \iff \|x_k-x_{\infty}\|_2 \to_{k\to \infty} 0 \end{align} That is, $(x_k)$ converges to $x_{\infty}$ in $d_1$ if and only if it converges to $x_{\infty}$ in $d_2$.
In fact, Riesz's theorem states that for a finite dimensional vector spaces, all norms are equivalent, so this is true for every distance defined by a norm in $\mathbb{R}^n$.