Where does this infinite series converge:

Sigma (k = 1 to infinity): 1/(9k^2 + 3k - 2)

term(k) = 1/[(3k+2)(3k-1)]

so S(1) = 1/(5 x 2) = 1/10
S(2) = 1/10 + 1/(8 x 5) = 1/8 or 2/16
S(3) = 1/8 + 1/88 = 12/88 = 3/22
S(4) = 3/22 + 1/154 = 1/7 or 4/28

it appears we have a nice pattern here and
S(n) = n/(6n+4)

This type of question usually comes up with the topic of "induction".

You would now have to prove that this conjecture is true by induction

Thanks Reiny! I couldn't factor that polynomial. Once there, the rest follows the book example

Each term in the sequence: 1/(9k^2 + 3k - 2) = 1/((3k+2)(3k-1)) = 1/3*(1/(3k-1) - 1/(3k+2))
When you take a sum of those terms, it's a telescoping series where the -1/(3k+2) cancels the next +1/(3k-1) term.
The sum of a partial series from terms 1 to n = 1/3*(1/2 - 1/(3n + 2)). Limit to infinity = 1/6

for your additional information:

You might want to remember that in these kind of telescoping series, the two factors are of this pattern:
(mk + a)(mk + b) where a - b = m

in our case they were (3k+2)(3k-1)
notice 2-(-1) = 3

Also notice that if you simplify your

1/3*(1/2 - 1/(3n + 2)) you get my n/(6n+4)

and Limit n/(6n+4) as n ---> ∞ = 1/6

5_13+7_61=11_20

To determine where the given infinite series converges, we can use the concept of convergence of series. Specifically, we will use the Ratio Test.

The Ratio Test states that for a series Σ(aₙ) to converge, the limit of the absolute value of the ratio of consecutive terms, lim(n→∞) |aₙ₊₁ / aₙ|, must be less than 1.

In the given series, aₙ = 1/(9k^2 + 3k - 2). We need to find the limit of the absolute value of the ratio |aₙ₊₁ / aₙ| as n approaches infinity.

Let's compute the ratio:

|aₙ₊₁ / aₙ| = |(1/(9(k+1)^2 + 3(k+1) - 2)) / (1/(9k^2 + 3k - 2))|
= |(9k^2 + 21k + 9)/(9k^2 + 27k + 20)|

Simplifying the expression further:

|aₙ₊₁ / aₙ| = | [(3k + 1)(3k + 9)] / [(k + 4)(3k + 5)] |
= |(3k + 1)(3k + 9)| / |(k + 4)(3k + 5)|

Now, we want to find the limit of this expression as k approaches infinity.

As k approaches infinity, the highest power term dominates in each factor. We can ignore the lower-degree terms:

|aₙ₊₁ / aₙ| = |3k * 3k| / |k * 3k|
= | 9k^2 / 3k^2 |
= | 3 |

Taking the limit as k approaches infinity:

lim(k→∞) |aₙ₊₁ / aₙ| = lim(k→∞) 3 = 3

Since the limit is 3 (greater than 1), the Ratio Test tells us that the series diverges. Therefore, the given infinite series does not converge.

jko[j