IMC 2018 · Day 1 · P1

Note that the sum of a series with positive terms can be either finite or $+\infty$, so for such a series, “converges” is equivalent to “is finite”.

Proof for (1) $\Longrightarrow$ (2): By the AM-GM inequality, $$\sqrt{\frac{a_n}{b_n}} = \sqrt{\frac{a_n}{c_n} \cdot \frac{c_n}{b_n}} \le \frac12 \left( \frac{a_n}{c_n} + \frac{c_n}{b_n} \right),$$ so $$\sum_{n=1}^{\infty} \sqrt{\frac{a_n}{b_n}} \le \frac12 \sum_{n=1}^{\infty} \frac{a_n}{c_n} + \frac12 \sum_{n=1}^{\infty} \frac{c_n}{b_n} < +\infty.$$ Hence, $\displaystyle\sum_{n=1}^{\infty} \sqrt{\frac{a_n}{b_n}}$ is finite and therefore convergent.

Proof for (2) $\Longrightarrow$ (1): Choose $c_n = \sqrt{a_n b_n}$. Then $$\frac{a_n}{c_n} = \frac{c_n}{b_n} = \sqrt{\frac{a_n}{b_n}}.$$ By the condition $\displaystyle\sum_{n=1}^{\infty} \sqrt{\frac{a_n}{b_n}}$ converges, therefore $\displaystyle\sum_{n=1}^{\infty} \frac{a_n}{c_n}$ and $\displaystyle\sum_{n=1}^{\infty} \frac{c_n}{b_n}$ converge, too.