Sequences & Series
Recurrence — |Σ(aₙ − 2/n²)|
nta_pyq_2026_jan
Grade 11
Question:
Let $a_1=1$ and for $n\geq1$, $a_{n+1}=\dfrac{1}{2}a_n+\dfrac{n^2-2n-1}{n^2(n+1)^2}$. Then $\left|\displaystyle\sum_{n=1}^\infty\left(a_n-\dfrac{2}{n^2}\right)\right|$ is equal to _____.
Step-by-Step Solution
Key Concept: Let $b_n=a_n-2/n^2$. Then $b_{n+1}=a_{n+1}-2/(n+1)^2=(a_n/2-1/n^2)=(b_n+2/n^2)/2-1/n^2=b_n/2$. So $\{b_n\}$ is a GP with ratio $1/2$.
$\left|\sum_{n=1}^\infty\left(a_n-\tfrac{2}{n^2}\right)\right|=2$.
Correct Answer: 2