2
$\begingroup$

TRUE/FALSE:

Let $a_n\geq0$ for each positive integer $n$. If the series $\sum\limits_{n=1}^{\infty}\sqrt{a_n}$ converges, then so does the series $\sum\limits_{n=1}^{\infty}\dfrac{a_n}{n^{1/4}}$

I tried to use Cauchy Schwarz but it is not coming.

$\endgroup$
2
  • 2
    $\begingroup$ If $a_n\le1$ then $\sqrt{a_n}\ge a_n\ge a_n/n^{1/4}$. $\endgroup$ Jan 18, 2023 at 9:24
  • 2
    $\begingroup$ For large $n$, it's necessarily the case that $a_n \leq \sqrt {a_n}$ and $n^{-1/4} \lt 1$, so the comparison test works. $\endgroup$ Jan 18, 2023 at 9:24

1 Answer 1

8
$\begingroup$

$\sqrt {a_n} \to 0$ so $a_n<1$ for $n$ sufficiently large. So $\frac {a_n} {n^{1/4}}\leq a_n \leq \sqrt {a_n}$. By Comparison Test $\sum \frac {a_n} {n^{1/4}}$ converges.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .