Calculus II - Series Comparison Test Problems

In summary, the conversation discusses two similar problems involving infinite series and the use of the direct or limit comparison test to determine their convergence or divergence. The speaker is struggling to find a suitable value for bn and asks for help in finding a relevant value of p, considering that ln(n) grows slower than any power x^p for p>0.
  • #1
invasmani
1
0

Homework Statement



I have two relatively similar problems:

1.) Sigma n=1 to infinity ((ln n)^3 / n^2)
2.) Sigma n=1 to infinity (1 / sqrt(n) * (ln n)^4)

I'm to prove their convergence or divergence using either the direct comparison test or the limit comparison test.

I understand both comparison tests, however, I am very much stumped on how to determine a working bn for the problems.

The Attempt at a Solution



I've tried using a known p-series, such as 1/(n^2) for problem #1, which is convergent. However, that p-series is not greater than the an in this case.

I'd appreciate any sort of direction on this.
 
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  • #2
ln(n) grows more slowly than any power x^p for p>0. Can you suggest an interesting value of p that's relevant for 1) or 2)?
 
Last edited:

Related to Calculus II - Series Comparison Test Problems

1. What is the Series Comparison Test in Calculus II?

The Series Comparison Test is a method used to determine the convergence or divergence of a series by comparing it to another known series. It states that if the terms of a given series are always less than or equal to the terms of a known convergent series, then the given series must also converge. Similarly, if the terms of a given series are always greater than or equal to the terms of a known divergent series, then the given series must also diverge.

2. How do you use the Series Comparison Test?

To use the Series Comparison Test, you must first identify a known convergent or divergent series to compare your given series to. Then, you must show that the terms of your given series are always less than or equal to (or greater than or equal to) the terms of the known series. If this is true, then your given series must converge (or diverge) by the Series Comparison Test.

3. What is the difference between the Series Comparison Test and the Limit Comparison Test?

The Series Comparison Test compares a given series to a known series, while the Limit Comparison Test compares a given series to a known function. Additionally, in the Series Comparison Test, the known series must have all positive terms, while in the Limit Comparison Test, the known function must be positive for all values greater than some specific number. However, both tests are used to determine the convergence or divergence of a given series.

4. Can the Series Comparison Test be used to prove absolute convergence?

Yes, the Series Comparison Test can be used to prove absolute convergence. If the terms of a given series are less than or equal to the terms of a known absolutely convergent series, then the given series must also converge absolutely. This is because if a series converges absolutely, it must also converge.

5. Are there any limitations to the Series Comparison Test?

One limitation of the Series Comparison Test is that it can only be used for series with positive terms. If a series has both positive and negative terms, other convergence tests such as the Alternating Series Test must be used. Additionally, the known series used for comparison must also be convergent or divergent, otherwise the Series Comparison Test cannot be applied.

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