Question

How do you find sum of the convergent series?

Answer 1

This is the problem.

Answer 2

hi

Answer 3

By using partial fractions,\[\frac{1}{9n^2 + 3n - 2} = \frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right).\]\[\sum_{n = 1}^\infty\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\]which is a telescoping series, so the result is simpliy 1/2. Finally,
\[\sum_{n = 0}^\infty \frac{4}{9n^2 + 3n - 2} = \frac{4}{3} \frac{1}{3} = \frac{2}{3}.\]

Answer 4

so you have to separate the equation for partial fractions to find out the result is 1/2

Answer 5

but where does the 1/3 come from?

Answer 6

oh, that must be a mistake. should it be 1/2?

Answer 7

It was a typo, the last line should be\[\sum_{n = 1}^\infty \frac{4}{9n^2 + 3n - 2} = \frac{4}{3} \frac{1}{2} = \frac{2}{3}.\]

Answer 8

Okay. Thanks for the help! :D I will redo this problem to remember this.

Answer 9

You are very welcome :).