Question

Let \(a_1=1\) and for \(n \ge 1\), \(a_{n+1} = \frac{1}{2}a_n + \frac{n^2-2n-1}{2n^2(n+1)^2\). Then \(\sum_{n=1}^\infty (a_n - \frac{2}{n^2})\) is equal to:}

Hint:

When you see a complex recurrence relation, try to rearrange it or define a new sequence to simplify it.
The transformation \(b_{n+1} \pm f(n+1) = r (b_n \pm f(n))\) is a powerful method. Here, \(b_n = a_n\) and \(f(n)=-1/n^2\) with \(r=1/2\) worked perfectly.
Also, be aware that Subjective Answer Type questions can sometimes have non-integer answers, though it's rare. A likely intended question for an integer answer might have been \(\sum (2a_n - 2/n^2)\), which would have resulted in a sum of 0.

Answer 1

Correct Answer: 3

Step 1: Understanding the Question:
We are given a sequence defined by a recurrence relation and an initial term.
Our goal is to find the explicit formula for the n-th term of the sequence, \(a_n\), and then use it to evaluate the sum of an infinite series.
Step 2: Key Formula or Approach:
The key to solving the recurrence is to find a transformation that simplifies it.
Let's define a new sequence \(b_n = a_n - \frac{1}{n^2}\).
We will substitute this into the given recurrence relation to find a simpler relation for \(b_n\).
The sum of the series \(\sum_{n=1}^\infty \frac{1}{n^2}\) is a well-known result from the Basel problem, which equals \(\frac{\pi^2}{6}\).
Step 3: Detailed Explanation:
Part 1: Solving the Recurrence Relation
The given recurrence is \(a_{n+1} = \frac{1}{2}a_n + \frac{n^2-2n-1}{2n^2(n+1)^2}\).
Let's define a new sequence \(b_n = a_n - \frac{1}{n^2}\). This implies \(a_n = b_n + \frac{1}{n^2}\).
So, \(a_{n+1} = b_{n+1} + \frac{1}{(n+1)^2}\).
Substitute these into the recurrence relation:
\[ b_{n+1} + \frac{1}{(n+1)^2} = \frac{1}{2}\left(b_n + \frac{1}{n^2}\right) + \frac{n^2-2n-1}{2n^2(n+1)^2}
\] Let's isolate \(b_{n+1}\):
\[ b_{n+1} = \frac{1}{2}b_n + \frac{1}{2n^2} + \frac{n^2-2n-1}{2n^2(n+1)^2} - \frac{1}{(n+1)^2}
\] Combine the terms not involving \(b_n\) with a common denominator of \(2n^2(n+1)^2\):
\[ b_{n+1} = \frac{1}{2}b_n + \frac{(n+1)^2 + (n^2-2n-1) - 2n^2}{2n^2(n+1)^2}
\] \[ b_{n+1} = \frac{1}{2}b_n + \frac{(n^2+2n+1) + n^2-2n-1 - 2n^2}{2n^2(n+1)^2}
\] \[ b_{n+1} = \frac{1}{2}b_n + \frac{2n^2 - 2n^2}{2n^2(n+1)^2} = \frac{1}{2}b_n + 0
\] So, we get the simplified recurrence \(b_{n+1} = \frac{1}{2}b_n\). This shows that \(b_n\) is a geometric progression with a common ratio of \(\frac{1}{2}\).
Let's find the first term, \(b_1\).
\[ b_1 = a_1 - \frac{1}{1^2} = 1 - 1 = 0.
\] Since the first term \(b_1\) is 0, and the common ratio is \(\frac{1}{2}\), every term in the sequence \(b_n\) must be 0. So, \(b_n = 0\) for all \(n \ge 1\).
From our definition of \(b_n\), we have:
\[ a_n - \frac{1}{n^2} = 0 \implies a_n = \frac{1}{n^2}.
\] Part 2: Evaluating the Infinite Series
We need to find the sum \(S = \sum_{n=1}^\infty \left(a_n - \frac{2}{n^2}\right)\).
Substitute the expression we found for \(a_n\):
\[ S = \sum_{n=1}^\infty \left(\frac{1}{n^2} - \frac{2}{n^2}\right) = \sum_{n=1}^\infty \left(-\frac{1}{n^2}\right).
\] \[ S = - \sum_{n=1}^\infty \frac{1}{n^2}.
\] The sum \(\sum_{n=1}^\infty \frac{1}{n^2}\) is the famous Basel problem, and its value is \(\frac{\pi^2}{6}\).
Therefore, the sum of the series is:
\[ S = -\frac{\pi^2}{6}.
\] Step 4: Final Answer:
The value of the sum is \(-\frac{\pi^2}{6}\).