determine whether the series converges of diverges
the sum from n=1 to infinity of
1/(the square root of (n^3+1))
I said that through the comparision test (comparing to 1/the square root of (n^3) the series converges
is this true?
That is a valid way to do it. Yes, every term is smaller than 1/n^3
To determine whether the series converges or diverges, we can indeed use the comparison test. The comparison test states that if 0 ≤ aₙ ≤ bₙ for all n and ∑ bₙ converges, then ∑ aₙ also converges. Conversely, if 0 ≤ bₙ ≤ aₙ for all n and ∑ bₙ diverges, then ∑ aₙ also diverges.
In this case, we want to compare the given series ∑(1/√(n³ + 1)) with the series ∑(1/√n³). Since √(n³ + 1) > √(n³) for all n, it follows that 1/√(n³ + 1) < 1/√n³ for all n.
Now, we need to determine whether the series ∑(1/√n³) converges or diverges. We can simplify this series by bringing √n³ to the denominator:
∑(1/√n³) = ∑(1/n^(3/2)).
This series is known as the p-series with p = 3/2. The p-series converges if p > 1 and diverges if p ≤ 1.
Since p = 3/2 > 1, we conclude that the series ∑(1/√n³) converges.
Using the comparison test, we established that 0 ≤ 1/√(n³ + 1) < 1/√n³, where the series ∑(1/√n³) converges. Thus, we can conclude that the series ∑(1/√(n³ + 1)) also converges.