1
$\begingroup$

Let $\sum_{n=1}^{\infty}a_n$ be a series with positive terms that converges. Show that the following series also does $\sum_{n=1}^{\infty}\dfrac{\sqrt{a_n}}{n}$.

Of course, if $a_n$ converges then by the Divergence test we get that $a_n\to0$. So $\exists N, \forall n>N$, $a_n<1$. From this should result that $\sqrt{a_n}<a_n$. But is that true if $a_n<1$? (I mean $\dfrac{1}{\sqrt{2}}<\dfrac{1}{2}$ isn't true at all)

$\endgroup$
  • $\begingroup$ No, that's not true. You could use $ab\le{1\over2}(a^2+b^2)$ with $a=\sqrt{a_n}$ and $b=1/n$, and the Comparison Test. $\endgroup$ – David Mitra Nov 23 '16 at 7:55
1
$\begingroup$

One may recall the Cauchy-Schwarz inequality $$ \left(\sum_{n=1}^N u_n v_n\right)^2\leq \left(\sum_{n=1}^N u_n^2\right) \left(\sum_{n=1}^N v_n^2\right) $$ giving here $$ \left(\sum_{n=1}^N\dfrac{\sqrt{a_n}}{n}\right)^2\leq \left(\sum_{n=1}^N a_n\right) \left(\sum_{n=1}^N \frac1{n^2}\right) $$ then one may conclude easily.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

hint

Use Cauchy-Schwarz.

$$\left(\sum_{n=1}^{\infty}\frac{\sqrt{a_n}}{n}\right)^2 \leq \sum_{n=1}^{\infty}a_n \, \sum_{n=1}^{\infty}\frac{1}{n^2}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @OlivierOloa yes you did, I was typing so I didn't see it. $\endgroup$ – Anurag A Nov 23 '16 at 7:59
  • $\begingroup$ @OlivierOloa Your answer is 1 minute older than this one, so it's probably fair to assume that OP started typing it before your answer was even posted... $\endgroup$ – 5xum Nov 23 '16 at 7:59
  • $\begingroup$ @Anurag No problem ;) $\endgroup$ – Olivier Oloa Nov 23 '16 at 8:00
  • 1
    $\begingroup$ @TheOscillator Your answer is fine, why deleting it? $\endgroup$ – Olivier Oloa Nov 23 '16 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.