Proving the metrics $d_1(x,y) $ and $d_2(x,y) $ are equivalent Prove that the metrics $$ d_1(x,y) = \sum\limits_{n=1}^\infty \frac{|x_n - y_n|}{ n^{2}(1+|x_n - y_n|)} $$ and $$ d_2(x,y) = \sum\limits_{n=1}^\infty \frac{|x_n - y_n|}{2^{n}(1+|x_n - y_n|)} $$ are equivalent on $ \omega = \mathbb{R}^\mathbb{N} $, the set of all real sequences.
I think to prove it, I have to show that fixed $ x\in \mathbb{R}^\mathbb{N} $, $ \epsilon > 0 $ and $ \gamma = f(\epsilon) $
$$ d_1(x,y)\geq \epsilon \implies d_2(x,y)\geq\gamma \tag{1}$$
$$ d_2(x,y)\geq \epsilon \implies d_1(x,y)\geq\gamma \tag{2}$$
But I do not know how should I choose $\gamma$ as a form of $\epsilon$ so I can prove this.
 A: We have $d_2(x,y)\leq 2d_1(x,y)$ because $2^{-n}<2n^{-2}$ for all $n\in \Bbb N.$ So the topology $T_2$ generated by $d_2$ is a subset of the topology $T_1$ generated by $d_1.$ 
To show that $T_1\subset T_2$ it suffices to show that for $x\in \Bbb R^{\Bbb N}$ and for $\delta >0$ there exists $\delta'\in (0,\delta]$ and $\delta^*>0$  such that  $B_{d_2}(x,\delta^*)\subset B_{d_1}(x,\delta').$
We will repeatedly use the fact that $f(t)=t/(1+t)$ is an increasing function for $t\geq 0.$  
Let $S=\sum_{n=1}^{\infty}n^{-2}.$
Let $\delta'=\min (\delta, 1, 2S).$  (Ignoring the fact that $2S>1.$ We will need $\delta' /2S\leq 1.$)
Take $n_1\in \Bbb N$ large enough that $\sum_{n=1+n_1}^{\infty}n^{-2}< \delta' /2.$ 
Let $\delta^*=(1/4S) \cdot \delta' \cdot  2^{-n_1}.$
If $y\in B_{d_2}(x,\delta^*)$ and $j\leq n_1$ then $|x_j-y_j|<\delta' /2S.$ Because if not , then $$d_2(x,y)\geq 2^{-j}\frac {|x_j-y_j|}{1+|x_j-y_j|}\geq$$     $$\geq 2^{-j}\frac {\delta' /2S}{1+\delta' /2S}\geq 2^{-j}\frac {\delta' /2S}{2}\geq 2^{-n_1}\frac {\delta' /2S} {2}\geq \delta^*.$$ 
So for $y\in B_{d_2}(x,\delta^*)$ we have   $$d_1(x,y)< \sum_{n=1}^{n_1} n^{-2}\frac {\delta' /2S }{1+\delta' /2S}+\sum_{n=1+n_1}^{\infty}n^{-2}<$$ $$<(\delta' /2 S)  S+\sum_{n=1+n_1}^{\infty}n^{-2}<\delta' /2+\delta' /2=\delta'.$$
