Answer
The series $\Sigma_{k=1}^{\infty} ca_k$ also diverges.
Work Step by Step
Here, we have
$\Sigma_{k=1}^{\infty} ca_k=\lim\limits_{n \to \infty} \Sigma_{k=1}^{n} ca_k$
or, $\Sigma_{k=1}^{\infty} ca_k=\lim\limits_{n \to \infty} c \Sigma_{k=1}^{n} a_k$
or, $\lim\limits_{n \to \infty} c \Sigma_{k=1}^{n} a_k = c \lim\limits_{n \to \infty} \Sigma_{k=1}^{n} a_k$
or, $\Sigma_{k=1}^{\infty} ca_k=(c) (\infty)$
or, $\Sigma_{k=1}^{\infty} ca_k=\infty$
Thus, we proved that the series $\Sigma_{k=1}^{\infty} ca_k$ also diverges.