Question

If \(a_1 = 1\) and for all \(n \ge 1\), \[ a_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2 (n+1)^2}, \] then the value of \[ \sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right) \] is equal to:

Hint:

For recurrence–series problems:
Try subtracting a known term to simplify the recurrence
Convert the sequence into a geometric progression if possible
Infinite GP sum formula: \( \dfrac{a}{1-r} \)

Answer 1

Correct Answer: 1

Step 1: Define a new sequence: \[ b_n = a_n - \frac{2}{n^2} \] Then, \[ b_{n+1} = a_{n+1} - \frac{2}{(n+1)^2} \] Substitute \(a_{n+1}\): \[ b_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2 (n+1)^2} - \frac{2}{(n+1)^2} \]
Step 2: Replace \(a_n = b_n + \dfrac{2}{n^2}\): \[ b_{n+1} = \frac{1}{2}b_n + \frac{1}{n^2} + \frac{n^2 - 2n - 1}{n^2 (n+1)^2} - \frac{2}{(n+1)^2} \] Simplifying the rational terms gives: \[ b_{n+1} = \frac{1}{2} b_n \]
Step 3: Compute the first term: \[ b_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1 \] Hence, \[ b_n = -\left(\frac{1}{2}\right)^{n-1} \]
Step 4: Evaluate the sum: \[ \sum_{n=1}^{\infty} b_n = -\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1} = -\frac{1}{1-\frac{1}{2}} = -1 \]