Solving Infinite Series: U(n) Convergence/ Divergence

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SUMMARY

The series U(n) = 1/(log(n))^(2*n) converges based on Cauchy's root test, which yields a limit of 0 as n approaches infinity, confirming convergence. Although the integral test was initially deemed applicable, it was skipped due to complexity. The discussion emphasizes that multiple convergence tests, including the comparison test, integral test, and ratio test, can be applied in any order, as long as a conclusive result is obtained. Ultimately, the order of tests does not affect the outcome, provided a valid test is used.

PREREQUISITES
  • Understanding of convergence tests in series, specifically Cauchy's root test and integral test.
  • Familiarity with logarithmic functions and their properties.
  • Knowledge of limits and their application in mathematical analysis.
  • Basic proficiency in calculus, particularly in evaluating series.
NEXT STEPS
  • Study the application of Cauchy's root test in various series.
  • Learn about the integral test for convergence and its practical applications.
  • Explore the comparison test and its limitations in determining series convergence.
  • Investigate the ratio test and its effectiveness in analyzing series behavior.
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in series convergence analysis will benefit from this discussion.

rohit dutta
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Given,
U(n)=1/(logn)^(2*n)

To find:
Whether the series ƩU(n) is convergent or divergent.

Sequence of tests to be followed:
*Comparison tests
*Integral tests
*D'Alembert's ratio test
*Raabe's test
*Logarithmic test
*Cauchy's root test

My approach:
Comparison test:
Since the series V(n) cannot be obtained from the given series, comparison test[which says Lt U(n)/V(n) as n→∞= some finite number (≠0)] fails. So, we must proceed.

Integral test:
Since, U(n) is a decreasing function of n, this test holds. One can perform the integral to find out whether the series is convergent or divergent. Say this integral is time taking. So, I decide to skip this test.

Ratio test:
U(n)/U(n+1)= [{log(n+1)}^2*(n+1)]/[{log(n)}^2*n]. I skip this too due to it's complexity.

Now, what I do is that I skip the intermediate tests and directly come to Cauchy's test.

Cauchy's root test:

[U(n)]^(1/n)= 1/[log(n)] (On solving)

Lt [U(n)]^(1/n) as n→∞= Lt 1/[log(n)] as n→∞= 0(<1). So, according to the test, the series converges.

Now, I initially said that the integral test holds but I decided to skip the test saying it was tough to integrate. Then, I moved on to the ratio test and came up with a complicated expression and once again, I decided to skip the test. I skipped the intermediate steps too and finally I felt that Cauchy's test was user-friendly :wink: and used it to get my answer.

My doubt is, can I perform any test that succeeds the previous one such that former test holds? What I mean to say is can I perform a test (say any test) which comes after the integral test to simplify the problem? The integral test did hold but it was tough whereas the Cauchy's test provided a simple approach. Will my approach always assure the right answer?

To be even more clear, will the result obtained by the integral test match the result obtained by Cauchy's test?

Also, do I need to follow the exact sequence of tests mentioned above? I believe that for a particular series more than one tests can hold so I always go in a random sequence depending on my analysis of the series. The textbook goes in this sequence so I wanted to clarify.
 
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You can use any of the tests in any order you like. The series either converges or diverges, and any test that gives you a conclusive result will give you the right answer. The answer doesn't change depending on which tests you perform in what order.

In this case, the root test has given you a conclusive answer: the series converges. You don't need to perform any other tests if you get a conclusive answer from one of them. The other tests may or may not give conclusive answers. If they do, they will all give the same answer. In general, use whichever test is easiest to apply to the given series.
 

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