Answer
(a) The integral test is applicable; the series diverges.
(b) The integral test is applicable; the series converges.
Work Step by Step
(a) Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{5x + 2}}$.
Notice that the series has positive terms. Since $f$ is decreasing and continuous for all $x \ge 1$, the integral test is applicable.
Evaluate:
$\mathop \smallint \limits_1^\infty \dfrac{1}{{5x + 2}}{\rm{d}}x = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{5}\left[ {\ln \left( {5x + 2} \right)} \right]_1^p = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{5}\left[ {\ln \left( {5p + 2} \right) - \ln 7} \right] = \infty $
Since $\mathop \smallint \limits_1^\infty \dfrac{1}{{5x + 2}}{\rm{d}}x$ diverges, by Theorem 9.4.4, the series also diverges.
(b) Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{1 + 9{x^2}}}$.
Since the series has positive terms and $f$ is decreasing and continuous for all $x \ge 1$, the integral test is applicable.
Evaluate $\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x$.
From Eq. (21) of Section 6.7, we know that
$\mathop \smallint \limits_{}^{} \dfrac{1}{{1 + {u^2}}}{\rm{d}}u = {\tan ^{ - 1}}u + C$
Therefore,
$\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{3}\left( {{{\tan }^{ - 1}}3x} \right)_1^p = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{3}\left( {{{\tan }^{ - 1}}3p - {{\tan }^{ - 1}}3} \right) = \dfrac{1}{3}\left( {\dfrac{\pi }{2} - {{\tan }^{ - 1}}3} \right)$
$\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x = \dfrac{\pi }{6} - \dfrac{1}{3}{\tan ^{ - 1}}3$
Since $\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x$ converges, by Theorem 9.4.4, the series also converges.