Question

AP Calculus help #5 convergence intervals

Answer 1

The given series is geometric, so it converges if

$\left|\dfrac2{x^2+1}\right| < 1$

For all real $x$, the denominator is always positive, so

$\left|\dfrac2{x^2+1}\right| = \dfrac2{x^2+1} < 1 \\\\ \implies \dfrac{x^2+1}2 > 1 \\\\ \implies x^2+1 > 2 \\\\ \implies x^2 > 1 \\\\ \implies x>1 \text{ or } x < -1$

When $x=\pm1$, the series diverges:

$\displaystyle \sum_{n=1}^\infty \left(\frac2{(\pm1)^2+1}\right)^n = \sum_{n=1}^\infty 1 \to \infty$

so (D) is the correct answer.