= Solution
Suppose first that $\|x_n-x_\infty\|\to0$. The reverse <triangle inequality> gives
$$
\big|\|x_n\|-\|x_\infty\|\big|
\leq\|x_n-x_\infty\|\longrightarrow0,
$$
so (i) implies (ii).
It also implies (iii). Given $\epsilon>0$, choose $N$ so that
$$
\|x_n-x_\infty\|^2<\frac\epsilon8
\qquad(n\geq N).
$$
By the <Parseval identity for a Hilbertian basis>, choose $I_0$ so that the basis tail of $x_\infty$ beyond $I_0$ is less than $\epsilon/8$. For $n\geq N$ and $I\geq I_0$,
$$
\sum_{i\geq I}|\langle x_n,e_i\rangle|^2
\leq
2\|x_n-x_\infty\|^2
+2\sum_{i\geq I}|\langle x_\infty,e_i\rangle|^2
<\frac\epsilon2.
$$
There are only finitely many $n<N$, so increasing $I$ makes each of their tails less than $\epsilon$. Thus one $I$ works for every $n$, proving (iii).
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