Let \( a_1 = 1 \) and for \( n \ge 1 \),
\[
a_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{n^2 (n+1)^2}.
\]
Then
\[
\sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right)
\]
is equal to
Show 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.
Step 1: Understanding the Question:
We are given a sequence defined by a recurrence relation along with an initial term.
Our objective is to determine the explicit formula for the sequence \(a_n\) and then use it to evaluate a related infinite series.
Step 2: Key Idea:
To simplify the recurrence relation, we introduce a transformed sequence:
\[
b_n = a_n - \frac{1}{n^2}
\]
This transformation helps reduce the recurrence to a simpler form.
We will also use the well-known result from the Basel problem:
\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
\]
Step 3: Detailed Solution: Part 1: Solving the Recurrence Relation
The given recurrence relation is:
\[
a_{n+1} = \frac{1}{2}a_n + \frac{n^2 - 2n - 1}{2n^2 (n+1)^2}
\]
Using the transformation \(a_n = b_n + \frac{1}{n^2}\), we write:
\[
a_{n+1} = b_{n+1} + \frac{1}{(n+1)^2}
\]
Substituting into the recurrence:
\[
b_{n+1} + \frac{1}{(n+1)^2}
= \frac{1}{2}\left(b_n + \frac{1}{n^2}\right)
+ \frac{n^2 - 2n - 1}{2n^2 (n+1)^2}
\]
Rearranging:
\[
b_{n+1}
= \frac{1}{2}b_n
+ \frac{1}{2n^2}
+ \frac{n^2 - 2n - 1}{2n^2 (n+1)^2}
- \frac{1}{(n+1)^2}
\]
Combining the constant terms using the common denominator \(2n^2 (n+1)^2\):
\[
b_{n+1}
= \frac{1}{2}b_n
+ \frac{(n+1)^2 + (n^2 - 2n - 1) - 2n^2}{2n^2 (n+1)^2}
\]
Simplifying the numerator:
\[
(n^2 + 2n + 1) + n^2 - 2n - 1 - 2n^2 = 0
\]
Thus:
\[
b_{n+1} = \frac{1}{2}b_n
\]
Now, compute the first term:
\[
b_1 = a_1 - \frac{1}{1^2} = 1 - 1 = 0
\]
Since \(b_1 = 0\) and the common ratio is \(\frac{1}{2}\), we get:
\[
b_n = 0 \quad \text{for all } n \ge 1
\]
Therefore:
\[
a_n = \frac{1}{n^2}
\]
Part 2: Evaluating the Infinite Series
We are asked to evaluate:
\[
S = \sum_{n=1}^{\infty} \left(a_n - \frac{2}{n^2}\right)
\]
Substitute \(a_n = \frac{1}{n^2}\):
\[
S = \sum_{n=1}^{\infty} \left(\frac{1}{n^2} - \frac{2}{n^2}\right)
= \sum_{n=1}^{\infty} \left(-\frac{1}{n^2}\right)
\]
So:
\[
S = -\sum_{n=1}^{\infty} \frac{1}{n^2}
= -\frac{\pi^2}{6}
\]
Step 4: Final Answer:
\[
\boxed{-\frac{\pi^2}{6}}
\]
