91Ó°ÊÓ

A series is said to be absolutely convergent if the series of the absolute values of its terms, \(\sum |a_n|\), is convergent. Mathematically, we write this as: if $$ \sum_{n=1}^{\infty} |a_n| < \infty $$ then the series is absolutely convergent.

A series is said to be convergent if the sequence of its partial sums, \(\sum a_n\), tends to a finite limit as \(n\) goes to infinity. Mathematically, we write this as: if $$ \lim_{n \to \infty} \sum_{k=1}^{n} a_k = L < \infty $$ then the series is convergent.

If a series is absolutely convergent, we can use the comparison test to show that it is also convergent. Let \(\sum a_n\) be an absolutely convergent series. Then, \(\sum |a_n|\) is convergent. By the comparison test, we know that if \(\sum |a_n|\) is convergent and \(0 \le a_n \le |a_n|\) for all \(n\), then \(\sum a_n\) is also convergent. First, note that for all \(n\), $$ 0 \le |a_n|, $$ which is always true. Now, let's consider the nonnegative part \(a_n^+ = \max(a_n,0)\) and the negative part \(a_n^- = \max(-a_n,0)\) of \(a_n\), for all \(n\). We have: $$ a_n = a_n^+ - a_n^- \quad \text{and} \quad |a_n| = a_n^+ + a_n^-. $$ Because \(\sum |a_n|\) converges, we must have \(\sum a_n^+\) and \(\sum a_n^-\) convergent. The sums \(\sum_{n=1}^{\infty} a_n^+\) and \(\sum_{n=1}^{\infty} a_n^-\) are convergent non-negative series. Therefore, they have finite limits: $$ \lim_{n \to \infty} \sum_{k=1}^{n} a_k^+ = L^+ < \infty \quad \text{and} \quad \lim_{n \to \infty} \sum_{k=1}^{n} a_k^- = L^- < \infty. $$

Since we have finite limits for both \(\sum a_n^+\) and \(\sum a_n^-\), we can calculate the limit for \(\sum a_n\): $$ \lim_{n \to \infty} \sum_{k=1}^{n} a_k = \lim_{n \to \infty} \sum_{k=1}^{n} (a_k^+ - a_k^-) = \lim_{n \to \infty} \left(\sum_{k=1}^{n} a_k^+ - \sum_{k=1}^{n} a_k^-\right) = L^+ - L^- = L < \infty. $$ Therefore, the series \(\sum a_n\) is convergent. By following the above step-by-step explanation, we have proved that if a series is absolutely convergent, then it must also be convergent.