Question

AP Calculus help #5 convergence intervals

Answer 1

The given series is geometric, so it converges if

`\left|\frac2{x^2+1}\right| < 1`

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

`\left|\frac2{x^2+1}\right| = \frac2{x^2+1} < 1 \\\\ \implies \frac{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.