Question

Match List-I with List-II and choose the correct option:

LIST-I (Infinite Series)	LIST-II (Nature of Series)
(A) \( 12 - 7 - 3 - 2 + 12 - 7 - 3 - 2 + \dots \)	(II) oscillatory
(B) \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \)	(IV) conditionally convergent
(C) \( \sum_{n=0}^{\infty} \left( (n^3+1)^{1/3} - n \right) \)	(I) convergent
(D) \( \sum_{n=1}^{\infty} \frac{1}{n \left( 1 + \frac{1}{n} \right)} \)	(III) divergent

Choose the correct answer from the options given below:

• A. A - I, B - II, C - III, D - IV
• B. A - II, B - I, C - III, D - IV
• C. A - II, B - IV, C - III, D - I
• D. A - II, B - IV, C - I, D - III

Hint:

When analyzing series, have a mental checklist: 1. Does the nth term go to zero? If not, it diverges. 2. Is it a known series (geometric, p-series, harmonic)? 3. Is it alternating? Use the Leibniz test. 4. Is it positive? Use comparison, limit comparison, ratio, or root tests.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question requires determining the nature (convergence, divergence, oscillation) of four different infinite series. We will use various tests for convergence. 

Step 2: Detailed Explanation: 
A. \( 12 - 7 - 3 - 2 + 12 - 7 - 3 - 2 + ... \) 
Let's look at the sequence of partial sums, \( S_k \). The repeating block of terms is \( 12, -7, -3, -2 \), and their sum is:

\[ 12 - 7 - 3 - 2 = 0 \]

Now, compute the partial sums: \[ S_1 = 12, \quad S_2 = 12 - 7 = 5, \quad S_3 = 5 - 3 = 2, \quad S_4 = 2 - 2 = 0, \quad S_5 = 0 + 12 = 12, ... \] The sequence of partial sums is \( 12, 5, 2, 0, 12, 5, 2, 0, ... \). Since this sequence does not approach a single limit but oscillates between a finite set of values, the series is oscillatory. 
Match: A - II 
B. \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \) 
This is the alternating harmonic series. By the Leibniz test for alternating series, since \( a_n = \frac{1}{n} \) is positive, decreasing, and \( \lim_{n \to \infty} a_n = 0 \), the series converges. To check for absolute convergence, we consider: \[ \sum |a_n| = \sum \frac{1}{n} \] which is the harmonic series and diverges. Since the series converges but not absolutely, it is conditionally convergent. 
Match: B - IV 
C. \( \sum_{n=0}^{\infty} \left( (n^3 + 1)^{1/3} - n \right) \) 
Let \( a_n = (n^3 + 1)^{1/3} - n \). We analyze the behavior of \( a_n \) for large \( n \): \[ a_n = n \left( 1 + \frac{1}{n^3} \right)^{1/3} - n \approx n \left( 1 + \frac{1}{3n^3} \right) - n = n + \frac{1}{3n^2} - n = \frac{1}{3n^2} \] By the Limit Comparison Test with \( b_n = \frac{1}{n^2} \), since \( \sum \frac{1}{n^2} \) is a convergent p-series (with \( p = 2 > 1 \)), the given series is convergent. 
Match: C - I 
D. \( \sum_{n=1}^{\infty} \frac{1}{n \left( 1 + \frac{1}{n} \right)} \) 
Let's simplify the general term: \[ a_n = \frac{1}{n+1} \] The series is: \[ \sum_{n=1}^{\infty} \frac{1}{n+1} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... \] This is the harmonic series, which is known to be divergent. 
Match: D - III 

Step 3: Final Answer: 
Combining the matches: A-II, B-IV, C-I, D-III. This corresponds to option (D).