Answer
(a) $\mathop \sum \limits_{n = 1}^\infty 2{a_n}$ converges.
(b) $\mathop \sum \limits_{n = 1}^\infty {3^n}{a_n}$ diverges
(c) $\mathop \sum \limits_{n = 1}^\infty \sqrt {{a_n}} $ converges.
Work Step by Step
(a) Consider $\mathop \sum \limits_{n = 1}^\infty 2{a_n}$.
Apply Theorem 2 (Root Test) of Section 11.5:
$L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{2{a_n}}}$
We get
$L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{2{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{2}\sqrt[n]{{{a_n}}} = \left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{2}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}}} \right)$
Since $\mathop {\lim }\limits_{n \to \infty } {2^{\dfrac{1}{n}}} = 1$ and $\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = \dfrac{1}{2}$, so $L = \dfrac{1}{2}$.
Because $L = \dfrac{1}{2} \lt 1$, by the Root Test: $\mathop \sum \limits_{n = 1}^\infty 2{a_n}$ converges.
(b) Consider $\mathop \sum \limits_{n = 1}^\infty {3^n}{a_n}$.
Apply Theorem 2 (Root Test) of Section 11.5:
$L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{3^n}{a_n}}}$
We get
$L = \mathop {\lim }\limits_{n \to \infty } \left( {3\sqrt[n]{{{a_n}}}} \right) = 3\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}}$
Since $\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = \dfrac{1}{2}$, so $L = \dfrac{3}{2}$.
Because $L = \dfrac{3}{2} \gt 1$, by the Root Test: $\mathop \sum \limits_{n = 1}^\infty {3^n}{a_n}$ diverges.
(c) Consider $\mathop \sum \limits_{n = 1}^\infty \sqrt {{a_n}} $.
Apply Theorem 2 (Root Test) of Section 11.5:
$L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}^{1/2}}}$
We get
$L = \mathop {\lim }\limits_{n \to \infty } {\left( {{a_n}^{\dfrac{1}{2}}} \right)^{\dfrac{1}{n}}} = {\left( {\mathop {\lim }\limits_{n \to \infty } {a^{\dfrac{1}{n}}}} \right)^{\dfrac{1}{2}}}$
Since $\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = \dfrac{1}{2}$, so $L = {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{2}}} \approx 0.7071$.
Because $L \approx 0.7071 \lt 1$, by the Root Test: $\mathop \sum \limits_{n = 1}^\infty \sqrt {{a_n}} $ converges.
