convergence answer and show all work

Question

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#3 Which of the following series converge? Choice 1 only, choice 2 only, choice 3 only, choice 1 and 2 only, choice 1 2 3 all

1) 1 is the lower limit ∞ is the upper limit function is ((1)/(n square root of n))

 2) 1 is the lower limit ∞ is the upper limit function is (1/(3^n)

 3) 1 is the lower limit ∞ is the upper limit function is (1)/(n ln n )

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#4 which of the following statements about the series (1 is the lower limit; ∞ is the upper limit ;function is (1)/(2^n-n)) is true

1. The series diverges by the nth term test
2. The series diverges by limit comparison to the harmonic series 1 is the lower limit ; ∞ is the upper limit ; function is (1)/(n)
3. The series converges by the nth term test
4. The series converges by limit comparison to the geometric series 1 is the lower limit ∞ is the upper limit function is (1)/(2^n)

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Answer 1

Let's analyze each series to determine if it converges or diverges:

#3:

1) Series with function (1 / (n * √n)):

This series can be compared to the p-series with p = 3/2. Since 3/2 is greater than 1, the p-series converges. Therefore, the given series also converges.

2) Series with function (1 / 3^n):

This is a geometric series with a common ratio (r) of 1/3. Geometric series converge if |r| < 1. In this case, |1/3| = 1/3, which is less than 1, so this series converges.

3) Series with function (1 / (n * ln(n)):

You can use the integral test to check convergence. If you integrate (1 / (x * ln(x))) with respect to x from 2 to ∞, the integral converges, indicating that the series also converges.

So, all three series (Choice 1, Choice 2, and Choice 3) converge.

#4:

The given series has a function (1 / (2^n - n)). We can use the limit comparison test to determine its convergence behavior.

We'll compare it to the series with the function (1 / 2^n):

(1 / (2^n - n)) / (1 / 2^n) = (2^n) / (2^n - n)

As n approaches infinity, the limit of this ratio is 1, which means the given series behaves similarly to the harmonic series. The harmonic series (1/n) diverges.

Therefore, the correct statement for #4 is "The series diverges by limit comparison to the harmonic series 1 is the lower limit; ∞ is the upper limit; the function is (1/n)."