Answer
$\Sigma(a_{n} +b_{n})$ will be divergent
Work Step by Step
Let us assume $\Sigma (a_{n} + b_{n})$ is convergent . It is given that $\Sigma a_{n}$ is convergent.
Theorem 8
$\to$ $\Sigma^{\infty}_{n=1} (a_{n} + b_{n}) = \Sigma^{\infty}_{n=1}a_{n} + \Sigma^{\infty}_{n=1}b_{n}$
By Theorem 8, we can say that $\Sigma(a_{n} + b_{n})-a_{n}=\Sigma b_{n}$ converges.
Which contradicts the given information, which says that $\Sigma b_{n}$ diverges.
Hence, $\Sigma(a_{n} +b_{n})$ will be divergent.
