Answer
$\Sigma_{k=1}^{\infty} (a_k \pm b_k) =\Sigma_{k=1}^{\infty} a_k \pm \Sigma_{k=1}^{\infty}b_k$; converges
Work Step by Step
Here, we have
$\Sigma_{k=1}^{\infty} (a_k \pm b_k)=\lim\limits_{n \to \infty} (a_k \pm b_k)$
or, $\lim\limits_{n \to \infty} (a_k \pm b_k)=\lim\limits_{n \to \infty} a_k \pm \lim\limits_{n \to \infty} b_k$
or, $\Sigma_{k=1}^{\infty} (a_k \pm b_k) =\Sigma_{k=1}^{\infty} a_k \pm \Sigma_{k=1}^{\infty}b_k$
We can say that if the series $\Sigma_{k=1}^{\infty} (a_k \pm b_k)$, then the series $\Sigma_{k=1}^{\infty} a_k \pm \Sigma_{k=1}^{\infty}b_k$ converges as well.
