For what value of p does $\large\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^p}$ converge?

Question

geerky42 · Answer

My guess would be $0<p<\infty$

geerky42 · Answer

Am I correct?

agent0smith · Answer

Isn't that just the p-test? p-Series Convergence: http://www.math.com/tables/expansion/tests.htm if p>1 it converges.

geerky42 · Answer

Well, with $(-1)^n$, it's different story. Using Alternating series test, $\sum (-1)^na_n$ converges if $\lim_{n \rightarrow \infty}a_n = 0$

agent0smith · Answer

Yeah i was thinking use the alternating series test... too tired though, i should be asleep :P

geerky42 · Answer

$a_n = \dfrac{1}{n^p}$ so I'm sure it's 0 < p < ∞

geerky42 · Answer

Oh lol ok.

agent0smith · Answer

Yeah it should converge for 0 < p < inf then, since the terms are decreasing.

geerky42 · Answer

yeah

geerky42 · Answer

Just want to make sure, sorry. Thanks.