How do you use the direct comparison test to determine if #Sigma 1/(sqrt(n^3+1))# from #[1,oo)# is convergent or divergent?

Answer 1

Diverges.

#1/sqrt(n^3+1) > 1/sqrt(2n^3) = 1/sqrt(2) 1/n^(3/2)#

then

#1/sqrt(2) sum 1/n^(3/2) < sum 1/(sqrt(n^3+1))# and
# sum 1/n^(3/2)# diverges because #3/2 < 2# then
#sum 1/(sqrt(n^3+1))# diverges
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Answer 2

To use the direct comparison test for the series ( \sum_{n=1}^\infty \frac{1}{\sqrt{n^3+1}} ), we need to find a series ( \sum_{n=1}^\infty b_n ) that satisfies the following conditions:

  1. ( 0 \leq \frac{1}{\sqrt{n^3+1}} \leq b_n ) for all ( n ).
  2. The series ( \sum_{n=1}^\infty b_n ) is known to converge or diverge.

Since ( \frac{1}{\sqrt{n^3+1}} ) approaches 0 as ( n ) approaches infinity, we need to find a series ( \sum_{n=1}^\infty b_n ) such that ( b_n ) dominates ( \frac{1}{\sqrt{n^3+1}} ) and is easier to analyze.

One possible choice is ( b_n = \frac{1}{\sqrt{n^3}} = \frac{1}{n^\frac{3}{2}} ).

Now, we need to check whether ( \sum_{n=1}^\infty \frac{1}{n^\frac{3}{2}} ) converges or diverges. This series is a ( p )-series with ( p = \frac{3}{2} > 1 ), so it converges.

By the direct comparison test, since ( 0 \leq \frac{1}{\sqrt{n^3+1}} \leq \frac{1}{\sqrt{n^3}} = \frac{1}{n^\frac{3}{2}} ) and ( \sum_{n=1}^\infty \frac{1}{n^\frac{3}{2}} ) converges, the series ( \sum_{n=1}^\infty \frac{1}{\sqrt{n^3+1}} ) also converges.

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Answer 3

To use the direct comparison test to determine the convergence of the series Σ(1/√(n^3 + 1)) from n = 1 to infinity, we need to find another series that we know the convergence status of and can compare it to. In this case, we can compare it to the series Σ(1/n^(3/2)) from n = 1 to infinity.

By comparing the terms of the two series, we can see that for all n ≥ 1, 1/√(n^3 + 1) ≤ 1/√(n^3), since adding 1 to n^3 will always make the denominator larger, resulting in a smaller fraction.

Now, we know that Σ(1/n^(3/2)) from n = 1 to infinity is a convergent p-series with p = 3/2, which converges because 3/2 > 1.

Since 1/√(n^3 + 1) ≤ 1/√(n^3) for all n ≥ 1 and Σ(1/n^(3/2)) from n = 1 to infinity converges, by the direct comparison test, Σ(1/√(n^3 + 1)) from n = 1 to infinity also converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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