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.


1 Answer 1


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 $a$.

  • $\begingroup$ 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. $\endgroup$
    – xbh
    Jan 20, 2019 at 9:43
  • $\begingroup$ 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. $\endgroup$
    – Ingix
    Jan 20, 2019 at 10:20
  • $\begingroup$ @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. $\endgroup$ Jan 20, 2019 at 11:09
  • $\begingroup$ @xbh How would you go about showing (c) with the Cauchy-Schwarz inequality? $\endgroup$ Jan 20, 2019 at 11:11
  • $\begingroup$ @s0ulr3aper07 See my first comment. The first half is the application of C-S inequality. $\endgroup$
    – xbh
    Jan 20, 2019 at 11:14

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