= Solution
The <comparison test for series> states that if $0\leq a_n\leq b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. The partial sums of $\sum a_n$ are increasing and bounded above by a constant plus the bounded partial sums of $\sum b_n$, so the <monotone bounded sequence> theorem applies.
If $\sum x_n$ converges, then $x_n^2\leq x_n$ proves convergence of $\sum x_n^2$. Also $x_n\to0$, so eventually $x_n\leq1/2$ and $x_n/(1-x_n)\leq2x_n$, proving convergence of $\sum x_n/(1-x_n)$. The converse for the square series is false: $x_n=1/(n+1)$ has convergent square series but divergent original series. The converse for the fraction series is true because $x_n\leq x_n/(1-x_n)$.
If $(x_n)$ is positive and decreasing and $\sum x_n$ converges, put $m=\lfloor n/2\rfloor$. Then
$$
(n-m)x_n\leq\sum_{k=m+1}^{n}x_k\longrightarrow0,
$$
so \b[$nx_n\to0$]. The converse fails for $x_n=1/(n\log n)$ for $n\geq2$: it is decreasing and $nx_n\to0$, but the <integral test for convergence> shows that its series diverges.
Even convergence of $\sum x_n$ need not imply $(n\log n)x_n\to0$. Choose integers $N_j=\lceil e^{2^j}\rceil$ and set $x_n=(N_j\log N_j)^{-1}$ for $N_{j-1}<n\leq N_j$. This is decreasing, and the $j$th block contributes at most $1/\log N_j\leq2^{-j}$, so the series converges. At $n=N_j$, however, $(n\log n)x_n=1$.
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