Answer
(a) $\mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^2} + k + 3}}{{2{k^2} + 1}} \ne 0$; the series diverges.
(b) $\mathop {\lim }\limits_{k \to \infty } {\left( {1 + \dfrac{1}{k}} \right)^k} \ne 0$; the series diverges.
(c) The limit $\mathop {\lim }\limits_{k \to \infty } \cos k\pi $ does not exist; the series diverges.
(d) $\mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{k!}} = 0$; the series may either converge or diverge.
Work Step by Step
(a) Let ${a_k} = \dfrac{{{k^2} + k + 3}}{{2{k^2} + 1}}$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^2} + k + 3}}{{2{k^2} + 1}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{1 + \dfrac{1}{k} + \dfrac{3}{{{k^2}}}}}{{2 + \dfrac{1}{{{k^2}}}}} = \dfrac{1}{2}$
Since $\mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^2} + k + 3}}{{2{k^2} + 1}} \ne 0$, by the divergence test, the series diverges.
(b) Let ${a_k} = {\left( {1 + \dfrac{1}{k}} \right)^k}$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } {\left( {1 + \dfrac{1}{k}} \right)^k}$
It is known that $\mathop {\lim }\limits_{k \to \infty } {\left( {1 + \dfrac{1}{k}} \right)^k} = {\rm{e}}$. Thus, $\mathop {\lim }\limits_{k \to \infty } {\left( {1 + \dfrac{1}{k}} \right)^k} \ne 0$.
By the divergence test, the series diverges.
(c) Let ${a_k} = \cos k\pi $.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \cos k\pi $
The limit does not exist because it oscillates between $1$ and $-1$. Therefore, the series diverges.
(d) Let ${a_k} = \dfrac{1}{{k!}}$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{k!}} = 0$
According to the divergence test, the series may either converge or diverge.
