= Solution
For $u_k=a^{2k+1}/(2k+1)>0$, the consecutive ratio is
$$
\frac{u_{k+1}}{u_k}=a^2\frac{2k+1}{2k+3}\longrightarrow a^2.
$$
Thus the <ratio test> proves convergence when $0<a<1$ and divergence when $a>1$. At $a=1$, the <series> is $\sum_{k\geq0}1/(2k+1)$, which diverges by comparison with $\tfrac12\sum_{k\geq0}1/(k+1)$. Therefore
$$
\boxed{\text{convergent for }0<a<1;\qquad\text{divergent for }a\geq1.}
$$
All terms are nonnegative, so convergence here is also <absolute convergence>.
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