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`sum_(n=1)^oo 1/(9n^2+3n-2)` Find the sum of the convergent series.

`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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