Let a_{1}=1 and for n ge 1, a_{n+1} = frac{1} | vedclass

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(B) Given the recurrence relation: $a_{n+1} = \frac{1}{2}a_{n} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$.This can be rewritten as $a_{n+1} - \frac{1}{2}a_{

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$2$

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(B) Given the recurrence relation: $a_{n+1} = \frac{1}{2}a_{n} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$.This can be rewritten as $a_{n+1} - \frac{1}{2}a_{n} = \frac{2}{(n+1)^{2}} - \frac{1}{n^{2}}$.Let $b_{n} = a_{n} - \frac{2}{n^{2}}$. Then $a_{n} = b_{n} + \frac{2}{n^{2}}$.Substituting this into the recurrence:$b_{n+1} + \frac{2}{(n+1)^{2}} = \frac{1}{2}(b_{n} + \frac{2}{n^{2}}) + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$.$b_{n+1} = \frac{1}{2}b_{n} + \frac{1}{n^{2}} - \frac{2}{(n+1)^{2}} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$.Simplifying the right side:$b_{n+1} = \frac{1}{2}b_{n} + \frac{(n+1)^{2} - 2n^{2} + n^{2}-2n-1}{n^{2}(n+1)^{2}} = \frac{1}{2}b_{n} + \frac{n^{2}+2n+1-2n^{2}+n^{2}-2n-1}{n^{2}(n+1)^{2}} = \frac{1}{2}b_{n}$.Since $b_{1} = a_{1} - \frac{2}{1^{2}} = 1 - 2 = -1$, we have $b_{n} = b_{1} \cdot (\frac{1}{2})^{n-1} = -(\frac{1}{2})^{n-1}$.Thus, $\sum_{n=1}^{\infty} b_{n} = \sum_{n=1}^{\infty} -(\frac{1}{2})^{n-1} = -\frac{1}{1 - 1/2} = -2$.Therefore, $|\sum_{n=1}^{\infty} b_{n}| = |-2| = 2$.

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 $|\sum_{n=1}^{\infty}(a_{n}-\frac{2}{n^{2}})|$ is equal to ........... .

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