Question

If the sequence {a^2} converges then the sequence {a} converges true or false and why.

Answer 1

From logic you may also know:
\[\Large p \Rightarrow q \Longleftrightarrow \neg q \Rightarrow \neg p \]
So if {a} diverges, then a^2 diverges.

Answer 2

Sometimes showing that the contrapositive holds true is easier than showing that the implication holds.

Answer 3

\[ \lim_{n \to \infty }|a_n^2 - L|=\lim_{n \to \infty }|(|a_n| - \sqrt L)(|a_n| + \sqrt L)| = \lim_{n\to \infty }|(|a_n| + \sqrt L)| |(|a_n| - \sqrt L)| \\

\implies \lim_{n\to \infty } (|a_n| - \sqrt L)| < \epsilon/ \sqrt L \]

Answer 4

since the sequence converges absolutely, it must converge ..

Answer 5

Just for the sake of completeness, here is the contrapositive.

\[\Large \neg q \Rightarrow \neg p \]
So we assume that \(a_n\) diverges, as given by the implication of the contrapositive, this means that:
\[\Large \lim_{n \to \infty} a_n=\infty\]
now we can square both sides to optain:
\[\Large \lim_{n \to \infty} a_n^2=\infty^2=\infty \]
so we have shown that \(a_n^2\) diverges from our assumption that \(a_n\) diverges, therefore we can state:
\[\Large \neg q \Rightarrow \neg p \Longleftrightarrow p \Rightarrow q \]
which shows that if \(a_n^2\) converges, so does \(a_n\)

Answer 6

seems that this would work too ...