`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`

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