The given series is geometric, so it converges if
[tex]\left|\dfrac2{x^2+1}\right| < 1[/tex]
For all real [tex]x[/tex], the denominator is always positive, so
[tex]\left|\dfrac2{x^2+1}\right| = \dfrac2{x^2+1} < 1 \\\\ \implies \dfrac{x^2+1}2 > 1 \\\\ \implies x^2+1 > 2 \\\\ \implies x^2 > 1 \\\\ \implies x>1 \text{ or } x < -1[/tex]
When [tex]x=\pm1[/tex], the series diverges:
[tex]\displaystyle \sum_{n=1}^\infty \left(\frac2{(\pm1)^2+1}\right)^n = \sum_{n=1}^\infty 1 \to \infty[/tex]
so (D) is the correct answer.
