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 $f\left( x \right) = \dfrac{x}{{1 + {x^2}}}$. Notice that the series has positive terms.
From the figure attached, we see that $f$ is continuous and decreasing for $x \ge 2$. Thus, the integral test is applicable.
Evaluate $\mathop \smallint \limits_2^\infty \dfrac{x}{{1 + {x^2}}}{\rm{d}}x$.
$\mathop \smallint \limits_2^\infty \dfrac{x}{{1 + {x^2}}}{\rm{d}}x = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{2}\left[ {\ln \left( {1 + {x^2}} \right)} \right]_2^p = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{2}\left[ {\ln \left( {1 + {p^2}} \right) - \ln 5} \right] = \infty $
Since $\mathop \smallint \limits_2^\infty \dfrac{x}{{1 + {x^2}}}{\rm{d}}x$ diverges, by Theorem 9.4.4, the series also diverges.
(b) The series has positive terms. Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{{{\left( {4 + 2x} \right)}^{3/2}}}}$.
From the figure attached, we see that $f$ is continuous and decreasing for $x \ge 0$. Thus, the integral test is applicable.
Evaluate $\mathop \smallint \limits_0^\infty \dfrac{1}{{{{\left( {4 + 2x} \right)}^{3/2}}}}{\rm{d}}x$.
$\mathop \smallint \limits_0^\infty \dfrac{1}{{{{\left( {4 + 2x} \right)}^{3/2}}}}{\rm{d}}x = - \mathop {\lim }\limits_{p \to \infty } \left[ {\dfrac{1}{{{{\left( {4 + 2x} \right)}^{1/2}}}}} \right]_0^p = - \mathop {\lim }\limits_{p \to \infty } \left[ {\dfrac{1}{{{{\left( {4 + 2p} \right)}^{1/2}}}} - \dfrac{1}{2}} \right] = \dfrac{1}{2}$
Since $\mathop \smallint \limits_0^\infty \dfrac{1}{{{{\left( {4 + 2x} \right)}^{3/2}}}}{\rm{d}}x$ converges, by Theorem 9.4.4, the series also converges.
