Answer
(a) $\mathop \sum \limits_{n = 1}^\infty \left( {{a_n} + \dfrac{1}{{{n^2}}}} \right)$ converges.
(b) $\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^n}{a_n}$ converges.
(c) $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{1 + {a_n}^2}}$ diverges.
(d) $\mathop \sum \limits_{n = 1}^\infty \dfrac{{\left| {{a_n}} \right|}}{n}$ converges.
Work Step by Step
(a) Write $\mathop \sum \limits_{n = 1}^\infty \left( {{a_n} + \dfrac{1}{{{n^2}}}} \right) = \mathop \sum \limits_{n = 1}^\infty {a_n} + \mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{n^2}}}$.
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{n^2}}}$ converges because it is a $p$-series with $p = 2 \gt 1$. Since $\mathop \sum \limits_{n = 1}^\infty {a_n}$ is given to be convergent, by Theorem 1 (Linearity of Infinite Series) of Section 11.2: $\mathop \sum \limits_{n = 1}^\infty {a_n} + \mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{n^2}}}$ converges.
Hence, $\mathop \sum \limits_{n = 1}^\infty \left( {{a_n} + \dfrac{1}{{{n^2}}}} \right)$ converges.
(b) It is given that $\mathop \sum \limits_{n = 1}^\infty {a_n}$ is an absolutely convergent series. This implies that $\mathop \sum \limits_{n = 1}^\infty \left| {{a_n}} \right|$ is convergent.
Consider the series $\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^n}{a_n}$. If we take the absolute values of the terms, we get $\mathop \sum \limits_{n = 1}^\infty \left| {{a_n}} \right|$ which is convergent. By Theorem 1 of Section 11.4 (Absolute Convergence Implies Convergence): $\mathop \sum \limits_{n = 1}^\infty {\left( { - 1} \right)^n}{a_n}$ also converges.
(c) We are given $\mathop \sum \limits_{n = 1}^\infty {a_n}$, an absolutely convergent series. By Theorem 3 (the $n$th Term Divergence Test) of Section 11.2:
$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$
Knowing $\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$, we test $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{1 + {a_n}^2}}$ using Theorem 3 (the $n$th Term Divergence Test) of Section 11.2:
$\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{1 + {a_n}^2}} = 1$
Since $\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{1 + {a_n}^2}} \ne 0$, by the $n$th Term Divergence Test, the series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{1 + {a_n}^2}}$ diverges.
(d) Let ${c_n} = \dfrac{{\left| {{a_n}} \right|}}{n}$ and ${d_n} = \left| {{a_n}} \right|$. Both $\left\{ {{c_n}} \right\}$ and $\left\{ {{d_n}} \right\}$ are positive sequences.
We apply the Limit Comparison Test:
$L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{c_n}}}{{{d_n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left| {{a_n}} \right|}}{n}\cdot\dfrac{1}{{\left| {{a_n}} \right|}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n} = 0$
Since $L=0$ and $\sum {d_n} = \sum \left| {{a_n}} \right|$ converges, by the Limit Comparison Test, $\mathop \sum \limits_{n = 1}^\infty \dfrac{{\left| {{a_n}} \right|}}{n}$ converges.
