Sequence of sequences with sup-norm goes to 0 but 2-norm stay positive

Question

I am trying to show sup-norm is strictly weaker than 2-norm in sequence space. In $\Bbb R^n$, we have sup-norm equivalent to 2-norm since we are doing finite sum, $\sqrt{\sum^Na_n^2}\le\sqrt{\sum^N||a_n||_\infty^2}=\sqrt N||a_n||_\infty$. But, in sequence space, we have infinitely many terms, I think it is possible to keep the sup-norm very low but the 2-norm say positive. Therefore, we can find a sequence of sequences converges to 0 in supnorm but stay positive in 2-norm. But I got stuck in finding such an example.

Answer 1

Let $(x_n)_{n=1}^\infty$ be a sequence of sequences defined as follows \begin{align*} x_1 & = (1,0,0\cdots)\\ x_2 & = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,\cdots)\\ x_3 & = (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},0,\cdots)\\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\vdots\\ x_n & = (\underbrace{\frac{1}{\sqrt n},\frac{1}{\sqrt n},\cdots,\frac{1}{\sqrt n}}_{\text{first }n \:\text{terms}},0,0\cdots)\\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\vdots\\ \lvert\lvert x_n\rvert\rvert_\infty&=\frac{1}{\sqrt{n}} \:\:\text{but}\:\:\lvert\lvert x_n\rvert\rvert_2 =1 \end{align*}