= Solution
Put $\phi(t)=t/(1+t)$. Every summand is nonnegative and at most $2^{-i}$, so the series defining $d$ converges and $d\leq1$. Symmetry follows from each $d_i$. If $d(x,y)=0$, every nonnegative summand is zero, hence every $d_i(x_i,y_i)=0$ and $x_i=y_i$. The converse is immediate.
For the <triangle inequality>, first observe that for $a,b\geq0$,
$$
\phi(a+b)=\frac a{1+a+b}+\frac b{1+a+b}\leq\frac a{1+a}+\frac b{1+b}=\phi(a)+\phi(b).
$$
Monotonicity of $\phi$ and the triangle inequality for $d_i$ therefore give
$$
\phi(d_i(x_i,z_i))\leq\phi(d_i(x_i,y_i))+\phi(d_i(y_i,z_i)).
$$
Multiply by $2^{-i}$ and sum the convergent nonnegative series. This proves the required triangle inequality for $d$, so \b[$d$ is a <metric>]. The construction combines a <bounded metric transform> with a <countable product metric>.
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