Let a_1=1 and for n >= 1, a_ n+1\ =(1 / 2) a_n+fracn^2-2 n-1n^2(n+1)^2. Then |sum _n=1^infinity(a_n-frac2n^2)| is equal to

a_n+1-(1 / 2) a_n=fracn^2-2 n-1n^2(n+1)^2=frac2 n^2-(n+1)^3n^2(n+1)^2 => a_ n+1-(1 / 2) a_ n=frac2( n+1)^2-frac1 n^2 n=1 a_2-(1 / 2) a_1=frac22^2-frac11^2 2[ a_3-(1 / 2) a_2=frac23^2-frac12^2] 2^2[a_4-(1 / 2) a_3=frac24^2-frac13^2] ⋮ 2^n-2[a_n-(1 / 2) a_n-1=frac2n^2-frac1(n-1)^2] 2^n-1[a_n+1-(1 / 2) a_n=frac2(n+1)^2-frac1n^2] Adding \ a_ n+1=frac2( n+1)^2-frac12^ n \ => a_ n=frac2 n^2-frac12^ n-1 => |sum _n=1^infinity(a_n-frac2n^2)| \ => |sum _n=1^infinity-frac12^n-1|=2

Let a 1 1 and for n geq 1 a n 1 frac12 a n fracn 2 2 n ... | null (Mathematics)

a_{n+1}-\frac{1}{2} a_{n}=\frac{n^{2}-2 n-1}{n^{2}(n+1)^{2}}=\frac{2 n^{2}-(n+1)^{3}}{n^{2}(n+1)^{2}}\\

\Rightarrow \mathrm{a}_{\mathrm{n}+1}-\frac{1}{2} \mathrm{a}_{\mathrm{n}}=\frac{2}{(\mathrm{n}+1)^{2}}-\frac{1}{\mathrm{n}^{2}}\\

\mathrm{n}=1 \mathrm{a}_{2}-\frac{1}{2} \mathrm{a}_{1}=\frac{2}{2^{2}}-\frac{1}{1^{2}}\\

2\left[\mathrm{a}_{3}-\frac{1}{2} \mathrm{a}_{2}=\frac{2}{3^{2}}-\frac{1}{2^{2}}\right]\\

2^{2}\left[a_{4}-\frac{1}{2} a_{3}=\frac{2}{4^{2}}-\frac{1}{3^{2}}\right]\\

⋮

\\2^{n-2}\left[a_{n}-\frac{1}{2} a_{n-1}=\frac{2}{n^{2}}-\frac{1}{(n-1)^{2}}\right]\\

2^{n-1}\left[a_{n+1}-\frac{1}{2} a_{n}=\frac{2}{(n+1)^{2}}-\frac{1}{n^{2}}\right]\\

Adding

\\ \mathrm{a}_{\mathrm{n}+1}=\frac{2}{(\mathrm{n}+1)^{2}}-\frac{1}{2^{\mathrm{n}}} \\ \Rightarrow \mathrm{a}_{\mathrm{n}}=\frac{2}{\mathrm{n}^{2}}-\frac{1}{2^{\mathrm{n}-1}}\\

\Rightarrow\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right| \\ \Rightarrow\left|\sum_{n=1}^{\infty}-\frac{1}{2^{n-1}}\right|=2

Explanation

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