Question

Найти сумму ряда ∑ниж.n=1,верх.∞

Answer 1

$\\ \sum_{1}^{\infty}{8\over (n+1)(n+2)}=8\sum_{1}^{\infty}{1\over (n+1)(n+2)}\\\\ {A\over (n+1)}+{B\over(n+2)}={1\over(n+1)(n+2)}\\\\ An+2A+Bn+B=1\\ \begin{cases}A+B=0\\2A+B=1\end{cases}\\ B=-A, 2A-A=1, A=1, B=-1\\ 8\sum_{1}^{\infty} \left ( {1\over n+1}-{1\over n+2} \right )\\ S=a_1+a_2+...+a_{n-1}+a_n\\ a_1={1\over 2}-{1\over 3}, a_2={1\over 3}-{1\over 4}\\ a_{n-1}={1\over n}-{1\over n+1}, a_n={1\over n+1}-{1\over n+2}\\\\ 8\sum_{1}^{\infty} \left ( {1\over n+1}-{1\over n+2} \right )=8\lim_{x\rightarrow \infty}\left ( {1\over 2}-{1\over n+2} \right )=8*{1\over 2}=4$