Answer
False
Work Step by Step
In order to determine the correct statement, we need to consider $a_n=\dfrac{1}{n^2}$ and $b_n=\dfrac{1}{n}$
We can notice that $\dfrac{1}{n}\leq \dfrac{1}{n^2}$ because $\dfrac{1}{n^2} \gt 0$
Also, by the p-series test ($\Sigma_{n=1}^{\infty} \dfrac{1}{n^p}\implies p \gt 1;$ series converges), so the series $\Sigma a_n$ converges because $p=2 \gt 1$.
Therefore, we can conclude that the series $\Sigma a_n$ converges but the series $\Sigma b_n$ diverges.
Thus, the given statement is false.
