n^2-14n+48=(n-6)(n-8)
а дробь на простейшие дроби
2/(n^2-14n+48)=(1/(n-8))-(1/(n-6))
проверка
(1/(n-8))-(1/(n-6))=((n-6)-(n-8))/(n-6)(n-8)=2/(n^2-14n+48)
Находим n-yю частичную сумму:
S_(n)=∑^(n)_(9)((1/(k-8)) -(1/(k-6)))=
=(1- (1/3)) + ( (1/2) - (1/4)) + ( (1/3) - (1/5)) + ((1/4) - ( 1/6))+...
+ (1/(n-10))-(1/(n-8)) + (1/(n-9)) - (1/(n-7)) + (1/(n-8)) + (1/(n-6))
= 1 +(1/2) - (1/(n-7)) + (1/(n-6))
По определению
S=lim_(n→∞)S_(n)=3/2