What is the sum of this series?

Question

Determine whether the series:

Summation as n = 1 and goes up to infinity of (1 / (n ^ 2 + 5n + 6).

Answer 1

sigma [1/(n^2 +5n +6)]
= sigma [1/(n+2)(n+3)]
= sigma [1/(n+2) - 1/(n+3)]
= 1/(1+2) - 1/(infinity +3)
= 1/3 - nearly 0
= 1/3

Answer 2

To determine if the series is convergent or divergent, do a ratio test of the (n+1) th term to the nth term:

(n + 1) term = 1/[(n + 1)^2 + 5(n + 1) + 6]
= 1/(n^2 + 2n + 1 + 5n + 5 + 6)
= 1/ n^2 + 7n + 12)
= 1/(n + 4)(n+3)

nth term = 1/(n^2 + 5n + 6)
= 1/(n + 3)(n + 2)

Ratio of (n + 1)th term to nth term :
(since we have reciprocals, invert the denominator and bring it up to the numerator):
[(n + 3)(n + 2)] / [(n + 4)(n + 3)

Which simplifies to:

(n + 2)/(n + 4)

For any n, this is less than 1, therefore series converges. i.e. the series does NOT tend to infinity.

Answer 3

s = 1/(n^2 + 5n +6)

s = 1(n+3)(n+2)

s = (1/(n+2)) - (1/(n+3))

s = [(1/3) - (1/4)] - [(1/4)-(1/5)]- [(1/5) - (1/6)] ------------

s = 1/3 -[(1/4 - 1/4) +(1/5 - 1/5) + (1/6 - 1/6)---------(infinity-infinity)]

s = 1/3 -0

s = 1/3

Answer 4

write it as 1/(n+3)(n+2) = 1/(n+2) -1/(n+3)
Now sum up and simplify
1/3 - 1/4
1/4 -1/5
1/5-1/6

1/(n+2)-1/(n+3) = 1/3 -1/(n+2) ===>1/3

Answer 5

I plugged it into Maple, and Maple says that it's 1/2. I don't have any idea how to get that, though.

Answer 6

n~261/2