Posted by Sean on Wednesday, March 18, 2009 at 12:44am.
Where does this infinite series converge:
Sigma (k = 1 to infinity): 1/(9k^2 + 3k  2)

Math  Reiny, Wednesday, March 18, 2009 at 7:49am
term(k) = 1/[(3k+2)(3k1)]
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

Math  Sean, Wednesday, March 18, 2009 at 11:16am
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)(3k1)) = 1/3*(1/(3k1)  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/(3k1) term.
The sum of a partial series from terms 1 to n = 1/3*(1/2  1/(3n + 2)). Limit to infinity = 1/6

Math  Reiny, Wednesday, March 18, 2009 at 12:08pm
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)(3k1)
notice 2(1) = 3

Math  Reiny, Wednesday, March 18, 2009 at 12:14pm
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

Math  Anonymous, Wednesday, March 18, 2009 at 1:37pm
jko[j

Math  deirra, Wednesday, March 18, 2009 at 4:35pm
5_13+7_61=11_20
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