= Solution
By part (ii) and completeness of each factor, choose $x_i\in X_i$ with $x_i^{(n)}\to x_i$. Let $x=(x_i)$. Given $\varepsilon>0$, choose $N$ such that $\sum_{i>N}2^{-i}=2^{-N}<\varepsilon/2$. Coordinate convergence gives a single $n_0$ for which $d_i(x_i^{(n)},x_i)<\varepsilon/2$ for every $i\leq N$ and $n\geq n_0$. Since $\phi(t)\leq t$,
$$
d(x^{(n)},x)\leq\sum_{i\leq N}2^{-i}d_i(x_i^{(n)},x_i)+\sum_{i>N}2^{-i}<\varepsilon.
$$
Thus $x^{(n)}\to x$ in the product metric, proving \b[the product is complete]. The finite head uses coordinate convergence; the summable tail is controlled uniformly, regardless of how large the remaining original coordinate distances are. This is <completeness of a countable product metric>.
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