Is the norm $||x||:= \sum\limits_{i=1}^{\infty} 2^{-i}|x_i|$ equivalent to usual norm in $l_2$?

Question

Is the norm $$||x||:= \sum_{i=1}^{\infty} 2^{-i}|x_i|$$ equivalent to usual norm in $l_2$? I have already shown that it is a norm, I expect that it is equivalent, but I can not prove it.

Answer 1

No, it isn't. For each $n\in\mathbb{N}$ define $e_n=(0,...0,1,0,...)$, the $1$ is in coordinate number $n$. With the "usual" norm this is a unit vector. With your norm we have $||e_n||=2^{-n}$. So there is no way there exists a constant $C>0$ such that the usual norm is at most $C$ times your norm for all vectors.

Answer 2

Hint: Can you find vectors $u$ with (usual) norm $1$ but $\|u\| = 2^{-i}$?