Home Practice Questions AP Calculus BC Series Question No. 5

Answer is: option4

Solution:

I.The given series converges by the Alternating Series Test.

However, \[ \sum_{n=1}^{\infty}\frac{1}{2n+1} \ge \sum_{n=1}^{\infty}\frac{1}{3n} \] and \[ \sum_{n=1}^{\infty}\frac{1}{3n} \] is divergent.

Hence the given series is conditionally convergent.

II. The given series is not alternating, in spite of the \( (-1)^n \). The factor \( \cos n \) is sometimes positive and sometimes negative.

Since \( |\cos n| \le 1 \) for all \( n \), \[ \frac{|\cos n|}{3^n}\le\frac{1}{3^n} \]

Hence \[ \sum_{n=1}^{\infty}\frac{|\cos n|}{3^n} \le \sum_{n=1}^{\infty}\frac{1}{3^n} \] (a convergent geometric series).

Therefore the given series converges absolutely.

III. The given series converges by the Alternating Series Test.

We note that \[ \sum_{n=1}^{\infty}\frac{1}{\sqrt n} = \sum_{n=1}^{\infty}\frac{1}{n^{1/2}} \] is a divergent \(p\)-series.

Hence the given series is conditionally convergent.