Question

Find the sum of the convergent series. $$ \sum_{n=1}^{\infty} \frac{1}{9 n^{2}+3 n-2} $$

Short answer

The sum of the given series is \(\frac{2}{15}\).

Step-by-step solution

Partial Fraction Decomposition

We want to express \(\frac{1}{9n^2+3n-2}\) as a difference of two terms. First factorize the denominator, which gives us \(9n^2+3n-2 = (3n-1)(3n+2)\). Now, express the fraction as a sum of partial fractions: \(\frac{A}{3n-1} + \(\frac{B}{3n+2}\), where A and B are constants we want to determine.

Find the Constants

To find A and B, rearrange the fraction as follows: \(1 = A(3n+2) + B(3n-1)\). Now, choose suitable values for n to simplify the equation and find A and B. Let's choose n=1/3, this will give \(A = \frac{1}{3}\), and then choose n=-2/3, this will give \(B = \frac{-1}{5}\). Now we can rewrite the original expression in its partial fraction form: \(\frac{1}{9n^2+3n-2} = \frac{1}{3} \cdot \frac{1}{3n-1} - \frac{1}{5} \cdot \frac{1}{3n+2}\).

Apply Telescoping Summation

Rewrite the series summation using the partial fraction decomposition, and observe how terms cancel out, due to the 'telescoping' property. Extending the series, the sum becomes: \[\sum_{n=1}^{\infty} \frac{1}{3} \cdot \frac{1}{3n-1} - \frac{1}{5} \cdot \frac{1}{3n+2} = \frac{1}{3} - \frac{1}{5} + \frac{1}{6} - \frac{1}{8} + \frac{1}{9}...\], where most terms cancel out.

Calculate the Series Sum

After the cancellation of terms: \[\sum_{n=1}^{\infty} \frac{1}{9 n^{2}+3 n-2} = \frac{1}{3} - \frac{1}{5} = \frac{2}{15}\]