# Show that for each $p\geq1$, $\exists \text{ a sequence } a_n$ such that $\sum a_n$ converges and $\sum |a_n^p|$ diverges

I'm trying to come up with a sequence $(a_n)$ such that $\forall p\geq1$, $\exists \text{ a sequence } a_n$ such that $\sum_{n=1}^{\infty} a_n$ converges and $\sum_{n=1}^{\infty} |a_n^p|$ diverges. There was a hint saying to play around with the form $\frac{(-1)^n}{n^q}$, but I don't quite see how this helps.

By alternating series test, since $\lim \frac{1}{n^q} = 0$ and since $\frac{1}{n^q}$ is decreasing as n gets larger, we have convergence. So $\sum a_n$ converges.

For the $\sum a_n^p$ case, we have $\lim \frac{(-1)^n}{n^{pq}}$, but wouldn't this still converge since $p>1$?

• What if you take $q=1/(2p)$? – Clement C. Nov 11 '16 at 4:01
• Forgive me if I'm missing something but if $q = \frac{1}{2p}$, then for the case $\sum a_n^p$ we would have $\frac{(-1)^n}{n^{1/2}}$. Wouldn't this still converge under the alternating series test? – Nikitau Nov 11 '16 at 4:07
• Yes. You may want to check what the alternating series test says. But also, the series that needs to converge is without the power p, as per your question. The one tgat will diverge has the power p, and more crucially has absolute values. – Clement C. Nov 11 '16 at 4:09
• Oh wait. I think I see it now. With the absolute values, we will have $\frac{1}{\sqrt {n} }$. This is simply a p series with $p \leq 1$. So it diverges. Thanks! – Nikitau Nov 11 '16 at 4:14
• You're welcome. – Clement C. Nov 11 '16 at 4:14