# 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.

• (c): some inequalities might be useful. (d): just choose some $p > 1$ in $a_n= n^{-p}$ and give it a shot, because $\sum a_n$ would converge decently "slowly". – xbh Jan 20 at 9:08

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 $$a$$.

• I think he means $(\sum \sqrt {a_n}/n)^2\leqslant (\sum a_n) (\sum 1/n^2)$. I would use $2ab \leqslant a^2 +b^2$ which is basically your inequality. – xbh Jan 20 at 9:43
• Both inequalities seem to work, and if you don't already know the Cauchy-Schwarz inequaliy (your approach), the one by @Math_QED is the simpler one to see/verify. – Ingix Jan 20 at 10:20
• @Math_QED Thank you! Small question though: Is there a way to look at the general term and intuitively say whether it'll converge or diverge? In a test scenario I won't have the luxury of already knowing that (c) is correct. – s0ulr3aper07 Jan 20 at 11:09
• @xbh How would you go about showing (c) with the Cauchy-Schwarz inequality? – s0ulr3aper07 Jan 20 at 11:11
• @s0ulr3aper07 See my first comment. The first half is the application of C-S inequality. – xbh Jan 20 at 11:14