We know that $\displaystyle 0<\dfrac{3}{\sqrt{10n}} < \dfrac{3}{\sqrt{9n}} = \dfrac{1}{\sqrt{n}}$ for any $n\ge 1$. **Considering this fact, what does the direct comparison test say about $\displaystyle\sum\limits_{n=1}^{\infty }\dfrac{3}{\sqrt{10n}}$ ?** [[☃ radio 3]]