= Solution
At $x=1/4$ the terms are $c_n$. Their decay to zero alone is insufficient for convergence. In fact
$$
\frac{c_{n+1}}{c_n}=\frac{2n+1}{2n+2}
\geq\frac{n+1}{n+2},
$$
because the difference has numerator $n\geq0$. Since $c_1=1/2$, induction gives $c_n\geq1/(n+1)$. The comparison <harmonic series> diverges, so
$$
\boxed{x=1/4:\ \text{the series diverges to }+\infty.}
$$
At $x=-1/4$ the terms are $(-1)^nc_n$. Part (ii) gives positive decreasing magnitudes tending to zero, so the <alternating series test> applies. To see the convergence directly, let $S_N=\sum_{n=1}^N(-1)^nc_n$. Then
$$
S_{2k+2}-S_{2k}=-c_{2k+1}+c_{2k+2}<0,
$$
while $S_{2k}\geq-c_1$, obtained by grouping $S_{2k}=-c_1+(c_2-c_3)+\cdots+(c_{2k-2}-c_{2k-1})+c_{2k}$. Thus the even <partial sums> have a finite limit. The odd <partial sums> differ from them by $-c_{2k+1}\to0$, so the whole <sequence> of <partial sums> has the same limit. Absolute convergence would require convergence of $\sum c_n$, already disproved. Hence
$$
\boxed{x=-1/4:\ \text{the series converges conditionally}.}
$$
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