Mathematics & Numerical Analysis

Consider the counterexample:
Let $a_n = (-1)^n$ and $b_n = (-1)^n$.
Both $(a_n)$ and $(b_n)$ are oscillatory and therefore divergent.
However, their quotient $\frac{a_n}{b_n} = \frac{(-1)^n}{(-1)^n} = 1$ is a constant sequence converging to 1.
Their product $a_n b_n = (-1)^n \times (-1)^n = (-1)^{2n} = 1$ is also a constant sequence converging to 1.

A series $\sum a_n$ is said to be absolutely convergent if the series of absolute values $\sum |a_n|$ converges.

Taking the absolute values of the terms:

By the p-series test, $\sum_{n=1}^\infty \frac{1}{n^p}$ converges whenever $p \gt 1$. Here $p = 2 \gt 1$, so the series of absolute values converges.

According to the p-series test (hyperharmonic series $\sum_{n=1}^\infty \frac{1}{n^p}$):
The series converges if $p \gt 1$.
The series diverges if $p \le 1$.

Analysis of the two series:

• For $\sum_{n=1}^\infty \frac{1}{n^{1/2}}$, $p = 1/2 \le 1 \implies$ divergent.
• For $\sum_{n=1}^\infty \frac{1}{n^{3/2}}$, $p = 3/2 \gt 1 \implies$ convergent.

A foundational theorem of mathematical analysis states: Absolute convergence implies convergence.

That is, if the series of absolute values $\sum_{n=1}^\infty |a_n|$ converges, then the series $\sum_{n=1}^\infty a_n$ must necessarily converge.
Proof follows from Cauchy criterion: $|\sum_{k=m}^n a_k| \le \sum_{k=m}^n |a_k| \lt \epsilon$.