Answer
Please see the explanations below.
Work Step by Step
(a) For $k \ge 1$, the following inequality is valid:
$\dfrac{1}{{{3^k} + 5}} \lt \dfrac{1}{{{3^k}}}$
Thus,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{3^k} + 5}} \lt \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{3^k}}}$
The geometric on the right-hand side is convergent because the ratio $\left| r \right| = \dfrac{1}{3} \lt 1$. Therefore, by the Comparison Test, the smaller series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{3^k} + 5}}$ converges.
(b) Since $ - 1 \le \sin k \le 1$, it follows that $0 \le {\sin ^2}k \le 1$. Thus,
$\dfrac{{5{{\sin }^2}k}}{{k!}} \le \dfrac{5}{{k!}}$
$\mathop \sum \limits_{k = 1}^\infty \dfrac{{5{{\sin }^2}k}}{{k!}} \le 5\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k!}}$
1. Now, we show that the series on the right-hand side $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k!}}$ converges.
For $k \ge 1$, we have
$\dfrac{1}{{1\cdot 2\cdot 3\cdot\cdot\cdot\left( {k - 1} \right)k}} \lt \dfrac{1}{{{2^k}}}$
Thus,
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k!}} \lt \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{2^k}}}$
The geometric series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{{2^k}}}$ converges because the ratio is $\left| r \right| = \dfrac{1}{2} \lt 1$. Thus, by the comparison test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k!}}$ converges.
2. Back to our result earlier, namely
$\mathop \sum \limits_{k = 1}^\infty \dfrac{{5{{\sin }^2}k}}{{k!}} \le 5\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{k!}}$
The series on the right-hand side is a constant times a convergent series, hence converges. Therefore, by the Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{{5{{\sin }^2}k}}{{k!}}$ also converges.
