Answer:
Absolutely convergent
0.82
Step-by-step explanation:
Use the alternating series test.
lim(n→∞) (1/√n) = 0
1/√n > 1/√(n+1) for n > 0
It is absolutely convergent.
Use error estimation.
1/(n+1)² < 0.01
(n+1)² > 100
n+1 > 10
n > 9
n = 10
Sum the first 10 terms.
∑₁¹⁰ (-1)ⁿ⁻¹/n²
= 1/1² − 1/2² + 1/3² − 1/4² + 1/5² − 1/6² + 1/7² − 1/8² + 1/9² − 1/10²
≈ 0.82
