Question ID: #666
Let $a_1=1$ and for $n\ge1$, $a_{n+1}=\frac{1}{2}a_{n}+\frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}.$ Then $\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|$ is equal to
Solution:
Given the recurrence relation:
$$a_{n+1} = \frac{1}{2}a_n + \frac{n^2 – 2n – 1}{n^2(n+1)^2}$$
Let’s simplify the fractional part:
$$\frac{n^2 – 2n – 1}{n^2(n+1)^2} = \frac{2n^2 – (n^2 + 2n + 1)}{n^2(n+1)^2}$$
$$= \frac{2n^2 – (n+1)^2}{n^2(n+1)^2} = \frac{2}{(n+1)^2} – \frac{1}{n^2}$$
Substitute this back into the recurrence:
$$a_{n+1} = \frac{1}{2}a_n + \frac{2}{(n+1)^2} – \frac{1}{n^2}$$
$$a_{n+1} – \frac{2}{(n+1)^2} = \frac{1}{2}a_n – \frac{1}{n^2}$$
$$a_{n+1} – \frac{2}{(n+1)^2} = \frac{1}{2}\left(a_n – \frac{2}{n^2}\right)$$
Let $b_n = a_n – \frac{2}{n^2}$.
Then the relation becomes $b_{n+1} = \frac{1}{2}b_n$.
This indicates that $b_n$ is a Geometric Progression (G.P.) with common ratio $r = \frac{1}{2}$.
Calculate the first term $b_1$:
$$b_1 = a_1 – \frac{2}{1^2} = 1 – 2 = -1$$
We need to find the sum of the infinite series $\sum_{n=1}^{\infty} b_n$:
$$S = \frac{b_1}{1-r} = \frac{-1}{1 – 1/2} = \frac{-1}{1/2} = -2$$
The required value is the absolute magnitude:
$$|S| = |-2| = 2$$
Ans. 2
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