Discussion
Given a sequence of real numbers \((a_n)\), define
\[ b_n = \begin{cases} a_n, & \text{if } a_n \geq 0 \\ 0, & \text{if } a_n \lt 0 \end{cases} \]
and
\[ c_n = \begin{cases} a_n, & \text{if } a_n \lt 0 \\ 0, & \text{if } a_n \geq 0 \end{cases} \]
for each positive integer \(n\).
Which ONE of the following statements is TRUE?
Answer: C
The TRUE statement is: if \(\sum a_n\) converges but \(\sum |a_n|\) does not, then \(\sum b_n\) does not converge.
The sequence \(b_n\) keeps the non-negative terms and \(c_n\) the negative ones, so
\[ \begin{aligned} b_n &= \tfrac{1}{2}(|a_n| + a_n) \\ |a_n| &= 2b_n - a_n \end{aligned} \]
If \(\sum a_n\) and \(\sum b_n\) both converged, \(\sum |a_n|\) would converge too. So when \(\sum a_n\) converges conditionally, \(\sum b_n\) must diverge, which also rules out the option saying it converges. The same argument with \(c_n = \tfrac{1}{2}(a_n - |a_n|)\) shows \(\sum c_n\) diverges as well, so the last option fails for the conditionally convergent series \(\sum \frac{(-1)^{n+1}}{n}\).
The first option fails for \(a_n = -1\): \(\sum a_n\) diverges, yet every \(b_n = 0\), so \(\sum b_n\) converges.
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