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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`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`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial