# If {$|a_n|$} is divergent, can {$a_n$} be convergent? If {$a_n^2$} is convergent, can {$a_n$} be divergent?

Can anyone help me prove these problems?

If {$$|a_n|$$} is a divergent sequence, can {$$a_n$$} be a convergent sequence?

If {$$a_n^2$$} is a convergent sequence, can {$$a_n$$} be a divergent sequence?

• What does ** mean – Jakobian Oct 18 '18 at 14:42
• sorry, I haven't been very especifil, it means raised to the power of 2 – Lily island Oct 18 '18 at 14:49

If {|$$a_n$$|} is a divergent sequence, can {$$a_n$$} be a convergent sequence?

This is false. As by a theorem: if the sequence {$$a_n$$} converges to A, then {|$$a_n$$|} converges to |A|.

The contrapositive states: If {|$$a_n$$|} is divergent then {$$a_n$$} is divergent.

If {$$a_n^2$$} is a convergent sequence, could {$$a_n$$} be a divergent sequence?

Consider $$a_n=(-1)^n$$

the square converges while the actual sequence does not

For the first part, try to prove by contradiction. That is, assume that $$\{a_n\}$$ converges to some limit $$a$$ and show that $$\{|a_n|\}$$ converges. What will the limit of the latter sequence be?

For the second part, try to construct a sequence that repeatedly "hops" between two values, say $$-1$$ and $$1$$.