Q:

Show $\sum_{n=1}^\infty \frac {\sqrt a_n}{n}$ is always convergent if $\sum_{n=1}^\infty a_n$ is a convergent series of positive real numbers.

The question is as follows:
Let $\sum_{n=1}^\infty a_n$ be a convergent series of positive real numbers. Then which of the following statements(s) is (are) true?
(a) $\sum_{n=1}^\infty (a_n)^2$ is always convergent
(b) $\sum_{n=1}^\infty \sqrt a_n$ is always convergent
(c) $\sum_{n=1}^\infty \frac {\sqrt a_n}{n}$ is always convergent
(d) $\sum_{n=1}^\infty \frac {\sqrt a_n}{n^{1/4}}$ is always convergent 

As per the answer key, (a) and (c) are correct. So far I have been able to prove (a) and I have a counter example to refute (b). The real struggle has been to prove (c) and refute (d). I would really appreciate some help here. 

To refute (b) I considered $a_n = \frac {1}{n^2}$
My proof for (a) is provided below:
$\sum_{n=1}^\infty a_n$ is convergent 
$\Rightarrow \lim_{n\to\infty} a_n=0$
$\Rightarrow \exists \ n_0 \in \mathbb{N}$ such that $a_n <1 \ ,\forall \ n \geq n_0 $
$\Rightarrow \ (a_n)^2 \leq \ a_n \ ,\forall \ n \geq n_0 $
Now, using the Direct Comparison Test, we can say that $ \sum_{n=1}^\infty (a_n)^2$ always converges.

A:

Hint for $(c)$:
Calculate $(\sqrt{a_n} - 1/n)^2 \geq 0$
and use the comparison test.
Hint for $(d)$:
Try $a_n = n^{-a}$ for some suitable $a$.

