Suppose $\sum_{n=1}^{\infty}\sqrt{{a_{n}}/{n}}$ is convergent. Prove that $\sum_{n=1}^{\infty}a_{n}$ is also convergent. Let $\{a_{n}\}$ be a decreasing sequence of non-negative real numbers. Suppose $\sum_{n=1}^{\infty}\sqrt{\frac{a_{n}}{n}}$ is convergent. Prove that $\sum_{n=1}^{\infty}a_{n}$ is also convergent.
My thought is to use the direct comparison test but I think I'm struggling with showing that $a_{n}\leq\sqrt{\frac{a_{n}}{n}}$ $\forall$ n $\in\mathbb{N}$.
Any help would be great. Thank you!

By the Cauchy condensation test, $\sum_{n=1}^{\infty} \sqrt{a_n/n}$ converges if and only if
$$ \sum_{n=1}^{\infty} 2^n\sqrt{a_{2^n}/2^n} = \sum_{n=1}^{\infty} \sqrt{2^n a_{2^n}} $$
converges. Likewise, $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{n=1}^{\infty} 2^n a_{2^n}$ converges. Now the conclusion follows from the observation that, if $b_n \geq 0$ and $\sum_n b_n$ converges, then so does $\sum_n b_n^2$.
