= Solution
This is a <geometric series> with ratio $r=2\alpha/(\alpha^2+1)$. Since $\alpha^2+1>0$, the condition $|r|<1$ is equivalent to
$$
\alpha^2+1-2|\alpha|=(|\alpha|-1)^2>0.
$$
It holds for every real $\alpha$ except $\alpha=\pm1$. At $\alpha=1$ the terms are all one; at $\alpha=-1$ they alternate between $-1$ and $1$. Neither sequence of terms tends to zero. Thus
$$
\boxed{\text{convergent exactly for }\alpha\in\mathbb R\setminus\{-1,1\}.}
$$
All the convergent cases have <absolute convergence>, including $\alpha=0$, when every term is zero.
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