On $\mathbb{R}^\omega$ are these metrics equivalent $\sum_{n=1}^\infty \frac{1}{2^n}\min(1,|a_n-b_n|)$ vs with $\frac{|a_n-b_n|}{n^2(1+|a_n-b_n|)}$ How can I show that on $\mathbb{R}^\omega$ these two metrics are equivalent $\sum_{n=1}^\infty \frac{1}{2^n}\min(1,|a_n-b_n|)$ and $\sum_{n=1}^\infty \frac{1}{n^2}\frac{|a_n-b_n|}{1+|a_n-b_n|}$?
Actually I'm not even sure whether they're equivalent but my intuition is that since as $a_n$ and $b_n$ get closer $\min(1,|a_n-b_n|)$ and $\frac{|a_n-b_n|}{1+|a_n-b_n|}$ will be nearly the same, and I assume that the $\frac{1}{2^n}$ and $1/n^2$ terms wont make a difference. However I don't know how to proceed with the proof. I'd love any kind of hint!
 A: We have to show that
$$
 d_1(a, b) = \sum_{n=1}^\infty \frac{1}{2^n}\min(1,|a_n-b_n|)
$$
and
$$
 d_2(a, b) = \sum_{n=1}^\infty \frac{1}{n^2}\frac{|a_n-b_n|}{1+|a_n-b_n|}
$$
generate the same topology on the space $\Bbb R^\omega$ (a countable product of copies of $\Bbb R$, which can be identified with the set of all mappings from $\Bbb N$ to $\Bbb R$, i.e. the set of all real-valued sequences.)
It suffices to show that
$$ \tag 1
 \forall r > 0: \exists s > 0: \forall a, b \in \Bbb R^\omega :
 d_1(a, b) < s \implies d_2(a, b) < r
$$
and also the other way around
$$ \tag 2
 \forall r > 0: \exists s > 0: \forall a, b \in \Bbb R^\omega :
 d_2(a, b) < s \implies d_1(a, b) < r \, .
$$
The essential estimate is
$$ \tag 3
 \frac 12 \min (1, x) \le \frac{x}{1+x} \le \min(1, x)
$$
for all $x \ge 0$, which is easy to verify. It shows that for each $n$ the terms
$$
\frac{1}{2^n}\min(1,|a_n-b_n|) \quad \text{and} \quad \frac{1}{n^2}\frac{|a_n-b_n|}{1+|a_n-b_n|}
$$
become small together. In addition we can use that the series remainders become small independently of $a$ and $b$.
Now let us prove $(1)$. For given $r > 0$, there is an $N$ such that $\sum_{n=N}^\infty \frac{1}{n^2} < r/2$. Then, using the right estimate in $(3)$,
$$
 d_2(a, b) = \sum_{n=1}^N \frac{1}{n^2}\frac{|a_n-b_n|}{1+|a_n-b_n|} + \frac r 2\\
\le \sum_{n=1 }^N \frac{1}{n^2} \min(1, |a_n - b_n| + \frac r 2 \\
\le 2^N \sum_{n=1 }^N \frac{1}{2^n} \min(1, |a_n - b_n| + \frac r 2 \\
\le 2^N d_1(a, b) + \frac r 2
$$
so that
$$
 d_1(a, b) <  \frac{r}{2 \cdot 2^N} \implies d_2(a, b) < r \, .
$$
The proof of $(2)$ is a bit easier: From the left estimate in $(3)$ and $2^{n+1} \ge n^2$ we get that
$$
\frac{1}{2^n}\min(1,|a_n-b_n|) \le \frac{2}{n^2} \frac{2|a_n-b_n|}{1+|a_n-b_n|}
$$
and therefore
$$
 d_1(a, b) \le 4 d_2(a, b) \, .
$$

Remark: The metrics $d_1$ and $d_2$ are not strongly equivalent. If we choose $a_n = 0$, and $b_n= 1$ if $n=k$ and $b_n = 0$ otherwise then
$$ 
d_1(a, b) = \frac{1}{2^k} \, , \, d_2(a, b) = \frac{1}{k^2}
$$
which shows that there is no constant $\beta > 0$ with the property
$$
 \forall a, b \in \Bbb R^\omega: d_2(a, b) \le \beta d_1(a, b) \, .
$$
