# Divergence test by absolute value?

Is it okay to apply divergence test on a series $\sum a_n$ and show that this series diverges by showing that $|a_n| = \infty$?

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 divergence test only mentions I must show divergence of the sequence, not the absolute value of sequence. I guess a better way of putting this question is to ask does divergence of absolute value of a sequence imply the divergence of the original sequence?

• What do you mean by "divergence test"? There are many test for divergence of a series. Also recall that if $\sum a_n$ converges then $\lim a_n =0$. In other words, if the terms $a_n$ do not tend to $0$ then $\sum a_n$ diverges (note that the converse is not true). So stating that $\lim (-1)^n n^2 \neq 0$ suffices to conclude that the series diverges. – Reveillark Mar 28 '15 at 3:37
• Note that the converse of your 'question', however, is true: if $\sum a_n$ diverges, then $\sum |a_n|$ diverges. – Leponzo Mar 16 '16 at 23:23
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.