Solution
= Solution
{c}
\b[The implication is false.] Set
$$
a_n=\frac1{n\log(n+1)}.
$$
Then $na_n=1/\log(n+1)\to0$. For $n\geq2$, $\log(n+1)\leq\log(2n)\leq2\log n$, so $a_n\geq1/(2n\log n)$. The <integral test> gives
$$
\int_2^R\frac{dx}{x\log x}=\log\log R-\log\log2\longrightarrow\infty.
$$
Thus the <comparison test> shows that $\sum a_n$ diverges.