Determine the convergence of the following series using the nth-partial sum or geometric series formula.

The sum of n=1 to inifitiy 1/(9n^2+3n-2)

How do I start? I'm guessing I should factor out the denominator but whats after that?

It is close to a geometric series, so use that test: the ratio of sucessive terms.

ratio= (9n^2+3n-2)/(9(n+1)^2 + 3(n+1) + 2)
and show it is less than one.

if it is geometric what is the sum of n?

To determine the convergence of the given series, we can start by factoring out the common denominator of 9n^2 + 3n - 2.

The series can be rewritten as:

1/(9n^2 + 3n - 2)

Now, let's apply the nth-partial sum technique.

The nth-partial sum formula for a series of the form (1/(9n^2 + 3n - 2)) is:

Sn = 1/(9(1)^2 + 3(1) - 2) + 1/(9(2)^2 + 3(2) - 2) + ... + 1/(9n^2 + 3n - 2)

Next, we can simplify the expression inside the parentheses:

Sn = 1/(9n^2 + 3n - 2)

Since this series is not a geometric series, we cannot directly apply the geometric series formula. However, we can attempt to transform it into a geometric series by analyzing the ratio of successive terms.

Let's calculate the ratio of the successive terms, which is:

Ratio = (9n^2 + 3n - 2) / (9(n+1)^2 + 3(n+1) - 2)

To determine if the series is geometric, we need to show that this ratio is less than one in order to apply the geometric series formula.

Simplifying the expression, we have:

Ratio = (9n^2 + 3n - 2) / (9(n+1)^2 + 3(n+1) - 2)

Now, you can simplify the numerator and denominator further, if needed.

If the ratio is indeed less than one, then the series is a geometric series, and we can find the sum using the formula for the sum of an infinite geometric series:

Sn = a / (1 - r)

Where "a" is the first term of the series and "r" is the common ratio between consecutive terms.