Series Convergent Test

[Mathyes]

Please help give me some clue for this problem.

The Problem: Determine whether or not the following series converges. Be sure to name the test you use to make your decision and show it.

Given: \(\displaystyle \sum\limits_{n = 2}^\infty {\frac{1}{{n{{(\ln n)}^2}}}} \)

Looking at the possible tests that I know such as, geometric series, P-series, harmonic series, comparison test, ratio test, root test, alternating series, and integral test, I can only see that the integral test is probably most fit for this problem. But if I am correct to use the integral test, then am I suppose to use "the integration by part" for solving the integral part of it? I tried the "integration by part" but it did not seem right in my work regardless if I assign \(\displaystyle u = \frac{1}{x}\) and \(\displaystyle dv = \frac{1}{{{{(\ln n)}^2}}}\) or vice versa. Please tell me which test would lead me to determine the convergence or divergence of this series.

Thanks,

 

[lookagain]

The Problem: Determine whether or not the following series converges. Be sure to name the test you use to make your decision and show it.Given: \(\displaystyle \sum\limits_{n = 2}^\infty {\frac{1}{{n{{(\ln n)}^2}}}} \) Looking at the possible tests that I know such as, geometric series, P-series, harmonic series, comparison test, ratio test, root test, alternating series, and > > > integral test < < < , I can only see that the integral test is probably most fit for this problem. But if I am correct to use the integral test, then am I suppose to use "the integration by part" for solving the integral part of it? I tried the "integration by part" but it did not seem right in my work regardless if I assign \(\displaystyle u = \frac{1}{x}\) and \(\displaystyle dv = \frac{1}{{{{(\ln n)}^2}}}\) or vice versa. Please tell me which test would lead me to determine the convergence or divergence of this series.

\(\displaystyle \displaystyle\int_{2}^{\infty}\dfrac{dx}{x[ln(x)]^2}\)

The integration by parts method is not what you want here.

Do a u-substitution method.

Let u = ln(x)

The exponent on u is -2.

\(\displaystyle du = \bigg(\dfrac{1}{x}\bigg)dx \ \ \ or \ \ \ \dfrac{dx}{x}\)

Before you do the integration, the integrand would look like: \(\displaystyle \ u^{-2}du.\)


Can you take it from there?

 

[Mathyes]

Thank you for helping me,

Yes, I can take it from there.

This is what I come up with at the end:

\(\displaystyle {\lim _{t \to \infty }}(\frac{1}{{\ln 2}}) - {\lim _{t \to \infty }}(\frac{1}{{\ln t}}) = 1.4426950409\).

But I still need to check one more thing which is I am not sure if this number 1.4426950409 is a finite number. If it is a finite number then the series converges. (I have to go back to the definition of a finite number).

 

[DrPhil]

Thank you for helping me,

Yes, I can take it from there.

This is what I come up with at the end:

\(\displaystyle {\lim _{t \to \infty }}(\frac{1}{{\ln 2}}) - {\lim _{t \to \infty }}(\frac{1}{{\ln t}}) = 1.4426950409\).

But I still need to check one more thing which is I am not sure if this number 1.4426950409 is a finite number. If it is a finite number then the series converges. (I have to go back to the definition of a finite number).

The answer is an irrational number, but clearly finite. It is < 2, for instance. You should leave the answer in the "exact" form, rather than writing down meaningless significant figures:

..........\(\displaystyle \dfrac{1}{\ln 2}\)

 

[Mathyes]

Got it, so the series converges by the integral test.

Thank you for showing me,