Is it okay to apply divergence test on a series $\sum a_n$ and show that this series diverges by showing that $\lim \limits_{n \to \infty } |a_n| = \infty$?
If I have alternating sequence of $(-1^n) n^2$, can I say absolute value of this, $n^2$ goes to infinity therefore its series diverges by Divergence Test?
The definition of "$n$-th term Divergence Test" for the series $\sum a_n$ only mentions that I must show the divergence of the sequence $(a_n)$, not the divergence of the sequence $(|a_n|)$. I guess a better way of putting this question is to ask the following. Does the divergence of absolute value of the terms of a sequence imply the divergence of the original sequence?
Thanks in advance!
Best Answer
The series $\sum_{n \geq 1} \frac{(-1)^n}{n^2}$ converges because $(1/n^2)_{n \geq 1}$ is a sequence of positive terms that decreases to zero, c.f alternating series test.
If I understood right, the answer to your question is no: there exists series $\sum a_n$ which converge, but $\sum |a_n|$ diverges, for example the harmonic series $\sum (-1)^n/n$. These ones are called conditionally convergent. In fact, given such a series, you can reorder its terms so that the reordered series converge to whatever you want.