Answer:
converges
Explanation:
The given series is:
Σn=0 to infinity (-1)^n + / √(1n + 8)
To determine if the series converges or diverges, we can use the comparison test, which involves comparing the given series to a known convergent or divergent series.
Let's compare the given series to a known convergent series. We know that the series Σ1/n^p converges if p > 1. In this case, the series Σ1/n^(1/2) is known to converge since (1/2) > 1.
Let's rewrite the given series in the form Σ1/n^p to compare:
Σn=0 to infinity (-1)^n + / √(1n + 8)
= Σn=0 to infinity (-1)^n + / (1n + 8)^(1/2)
Now we can use the comparison test by comparing the given series to the convergent series Σ1/n^(1/2). Since the terms of the given series are positive (taking the absolute value of (-1)^n), we can ignore the negative sign.
|(-1)^n + / (1n + 8)^(1/2)| ≤ 1 / (1n + 8)^(1/2) (taking absolute values)
As n approaches infinity, 1/(1n + 8)^(1/2) approaches 0, and since 0 is less than 1, the given series is also smaller than the convergent series Σ1/n^(1/2).
Therefore, by the comparison test, the given series converges since it is smaller than the convergent series Σ1/n^(1/2). So the correct answer is "converges".