n^2+4n+3=(n+1)(n+3)
тогда
8/(n^2+4n+3)= A/(n+1)+ B/(n+3)
8=(A+B)n+(3A+B)
A+B=0
3A+B=8
2A=8
A=4
B=-4
S_(n)= ∑ ^(n)_(1)(4/(k+1) - 4/(k+3))=
=(4/2) - (4/4)+
+(4/3) - ( 4/5)+
+(4/4) - (4/6)+
+(4/5) - (4/7)+
...
+4/(n-1) - 4/(n+1)+
+(4/n) - 4/(n+2) +
+4/(n+1) - 4/(n+3)
S_(n)=(4/2) + (4/3) + ... - 4/(n+2) - 4/(n+3)
=(10/3)+ ... - 4/(n+2) - 4/(n+3)
По определению:
S=lim_(n → ∞ )S_(n)=10/3