Let $X$ be a Banach space, let $\{w_n\}_{n=1}^{\infty}$ be a sequence of (strictly) positive real numbers, and consider the two associated topological spaces $\prod_{n=1}^{\infty} X$ and $$ X_1:=\left\{ x \in \prod_{n=1}^{\infty} X:\, \sum_{n=1}^{\infty} w_n \|x\|_X < \infty \right\} $$ where $\|\cdot\|_X$ is the norm on $X$ and $X_1$ is equipped with the metric $$ d_1((x_n),(y_n)):= \sum_{n=1}^{\infty} w_n \|x_n - y_n\| . $$ Since $\prod_{n=1}^{\infty} X$ is a countable product then its topology is given by the metric $$ d_{\prod}((x_n),(y_n)):= \sum_{n=1}^{\infty} \frac{1}{2^n}\frac{\|x_n-y_n\|}{1 +\|x_n-y_n\|}. $$
It seems to me that there should be a constant $C>0$ such that $d_{\prod}\leq C d_1$ but not ther other way around. I.e.: The metric on $\prod_{n=1}^{\infty} X$ restricted to the subset $X_1$ should be strictly coarser than the topology on $X_1$ defined by the metric $d_1$. But is this true?
There need not be any $C$ such that $d_{\prod} \leq C d_1$. Take $X=\mathbb R$ and suppose $w_n=\frac1 {4^{n}}$. Let $k$ be such that $2^{k} >2C$. Let $(x_n)=(0,0,...)$ and $y_n=1$ when $n=k$, $y_n=0$ when $n \neq k$. Then the inequality becomes $\frac 1{2^{k+1}} \leq C \frac 1 {4^{k}}$ or $2^{k} \leq 2C$.
There may be no such inequality in the opposite direction either.