Are these metrics equivalents?

Question

Let $B(\mathbb{N}, \mathbb{R})$ be the set of all bounded sequences $x=(x_i)$ of real numbers. We would like to consider in $B(\mathbb{N}, \mathbb{R})$ two different metrics

1. $d_1(x,y)=\sum_{n=1}^{\infty}\dfrac{|x_n-y_n|}{2^n}$

2. $d_2(x,y)=\sum_{n=1}^{\infty}=\dfrac{|x_n-y_n|}{n^{\alpha}},~\alpha>1.$

Are these metrics in $B(\mathbb{N}, \mathbb{R})$ equivalent? Is pretty clear to me that there exist $C>0,$ such that $d_2\leq Cd_1$. But the opposite inequality is not so obvius....

Answer 1

No, they're not equivalent. Consider $x^k$ where $x^k_n = n^{\alpha} \unicode{x1D7D9}_{n \leq k}$. Then $$ d_1(0, x^k) = \sum_{n=1}^k \frac{n^{\alpha} }{2^n} \leq \sum_{n=1}^{\infty} \frac{n^{\alpha} }{2^n} = M< \infty$$ for some $M$, so $(x^k)$ is a bounded sequence w.r.t. $d_1$. However $$ d_2(0, x^k) = k$$ which diverges to $\infty$ as $k \to \infty$. Hence $(x^k)$ is unbounded w.r.t. $d_2$.