Question

The series \(\dfrac{1}{2} - \dfrac{2}{3}\cdot\dfrac{1}{2^3} + \dfrac{3}{4}\cdot\dfrac{1}{3^3} - \dfrac{4}{5}\cdot\dfrac{1}{4^3} + \cdots\) is:

• A. Conditionally convergent
• B. Absolutely convergent
• C. Divergent
• D. None of the above

Hint:

Write the general term as \((-1)^{n+1}\dfrac{n}{(n+1)n^3}\) and compare \(|a_n|\) with \(1/n^3\).

Answer 1

The Correct Option is B

Step 1: Identify the general term. The \(n\)-th term is \(a_n = (-1)^{n+1}\dfrac{n}{n+1}\cdot\dfrac{1}{n^3} = (-1)^{n+1}\dfrac{1}{n^2(n+1)}\). Check: at \(n=1\), \(a_1 = 1/2\); at \(n=2\), \(a_2 = -1/(4\cdot 3) = -1/12\), matching the given series.

Step 2: To test absolute convergence, look at \(|a_n| = \dfrac{1}{n^2(n+1)}\). Since \(n^2(n+1) = n^3+n^2 > n^3\) for every \(n \geq 1\), we get \(|a_n| < \dfrac{1}{n^3}\).

Step 3: The series \(\sum \dfrac{1}{n^3}\) is a \(p\)-series with \(p = 3 > 1\), so it converges. By the comparison test, \(\sum |a_n|\) also converges.

Step 4: Since \(\sum |a_n|\) converges, the series converges absolutely, not merely conditionally. So the answer is option (B), \(\boxed{\text{Absolutely convergent}}\).