 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 toIt holds for every real $\alpha$ except $\alpha =\pm 1$. At $\alpha =1$ the terms are all one; at $\alpha =-1$ they alternate between $-1$ and $1$. Neither sequence of terms tends to zero. ThusAll the convergent cases have absolute convergence, including $\alpha =0$, when every term is zero.