Answer:
a) -7/9
b) 16 / (n² + 15n + 56)
c) 1
Explanation:
When n = 1, there is only one term in the series, so a₁ = s₁.
a₁ = (1 − 8) / (1 + 8)
a₁ = -7/9
The sum of the first n terms is equal to the sum of the first n−1 terms plus the nth term.
sₙ = sₙ₋₁ + aₙ
(n − 8) / (n + 8) = (n − 1 − 8) / (n − 1 + 8) + aₙ
(n − 8) / (n + 8) = (n − 9) / (n + 7) + aₙ
aₙ = (n − 8) / (n + 8) − (n − 9) / (n + 7)
If you wish, you can simplify by finding the common denominator.
aₙ = [(n − 8) (n + 7) − (n − 9) (n + 8)] / [(n + 8) (n + 7)]
aₙ = [n² − n − 56 − (n² − n − 72)] / (n² + 15n + 56)
aₙ = 16 / (n² + 15n + 56)
The infinite sum is:
∑₁°° aₙ = lim(n→∞) sₙ
∑₁°° aₙ = lim(n→∞) (n − 8) / (n + 8)
∑₁°° aₙ = 1