Calculus of a Single Variable, Chapter 9, 9.2, Section 9.2, Problem 34

sum_(n=1)^oo1/(9n^2+3n-2)
Let's rewrite the n'th term of the sequence as,
a_n=1/(9n^2+3n-2)
=1/(9n^2+6n-3n-2)
=1/(3n(3n+2)-1(3n+2))
=1/((3n+2)(3n-1))
Now let's carry out partial fraction decomposition,
1/((3n+2)(3n-1))=A/(3n+2)+B/(3n-1)
Multiply the above equation by LCD,
1=A(3n-1)+B(3n+2)
1=3An-A+3Bn+2B
1=(3A+3B)n-A+2B
Equating the coefficients of the like terms,
3A+3B=0 -----------------(1)
-A+2B=1 ------------------(2)
From equation 1,
3A=-3B
A=-B
Substitute A in equation 2,
-(-B)+2B=1
B+2B=1
3B=1
B=1/3
A=-1/3
a_n=(-1/3)/(3n+2)+(1/3)/(3n-1)
a_n=1/(3(3n-1))-1/(3(3n+2))
Now we can write down the n'th partial sum of the series as:
S_n=(1/(3(3-1))-1/(3(3+2)))+(1/(3(3*2-1))-1/(3(3*2+2)))+..........+(1/(3(3n-1))-1/(3(3n+2)))
S_n=(1/6-1/15)+(1/15-1/24)+.........+(1/(3(3n-1))-1/(3(3n+2)))
S_n=(1/6-1/(3(3n+2)))
sum_(n=1)^oo1/(9n^2+3n-2)=lim_(n->oo)S_n
=lim_(n->oo)(1/6-1/(3(3n+2)))
=1/6