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If every factor is a complete metric space, the countable product metric is complete. For a product-Cauchy sequence, the inequality $2^{-i}\varphi (d_i(x_i^{(m)},x_i^{(n)}))\le D(x^{(m)},x^{(n)})$, with $\varphi (t)=t/(1+t)$, makes each coordinate sequence Cauchy. Its coordinate limits define a product point. Coordinatewise convergence and the uniformly small summable tail imply convergence to that point in $D$. It matters that the original distances are bounded before they are summed: summing arbitrary unbounded metrics directly need not be finite.

Put $\varphi (t)=t/(1+t)$. Every summand is nonnegative and at most $2^{-i}$, so the series defining $d$ converges and $d\le 1$. 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 $a,b\ge 0$,Monotonicity of $\varphi$ and the triangle inequality for $d_i$ therefore giveMultiply by $2^{-i}$ and sum the convergent nonnegative series. This proves the required triangle inequality for $d$, so $d$ is a metric. The construction combines a bounded metric transform with a countable product metric.