Prove relationships between a given series

by RoRi

March 29, 2016

Let $\{ a_n \}$ be a sequence of positive terms.

Prove or give a counterexample: if $\sum_{n=1}^{\infty} a_n$ diverges then $\sum_{n=1}^{\infty} a_n^2$ also diverges.
Prove or give a counterexample: if $\sum_{n=1}^{\infty} a_n^2$ converges then $\sum_{n=1}^{\infty} \frac{a_n}{n}$ also converges.

Counterexample. Let $a_n = \frac{1}{n}$ . Then
$\sum_{n=1}^{\infty} \frac{1}{n}$

diverges. Furthermore, $a_n^2 = \frac{1}{n^2}$ and

$\sum_{n=1}^{\infty} \frac{1}{n^2}$

converges.
Proof. We know from a previous exercise (Section 10.20, Exercise #51) we know
$\sum a_n \quad \text{converges} \quad \implies \quad \sum \sqrt{a_n}n^{-p} \quad \text{converges}$

for $p > \frac{1}{2}$ . So, for this case define $b_n = a_n^2$ . Then $\sum b_n = \sum a_n^2$ converges by hypothesis. Furthermore, (since $a_n > 0$ so $\sqrt{a_n^2} = a_n$ ),

$\sum \sqrt{b_n} n^{-1} = \sum \frac{a_n}{n}.$

Since this is the case $p = 1$ , we have proved the statement $. \qquad \blacksquare$

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply