Does the series 1/ln(n) converge or diverge?
Understand the Problem
The question is asking whether the series 1/ln(n) converges or diverges as n approaches infinity. To determine this, we can apply convergence tests such as the comparison test or the integral test.
Answer
Diverges
Answer for screen readers
The series ( \sum_{n=2}^{\infty} \frac{1}{\ln(n)} ) diverges.
Steps to Solve
- Identify the Series to Analyze
We are looking at the series given by the terms ( \frac{1}{\ln(n)} ) as ( n ) approaches infinity. We need to determine whether the series ( \sum_{n=2}^{\infty} \frac{1}{\ln(n)} ) converges or diverges.
- Choose a Test for Convergence
A suitable test for this series is the Integral Test. According to this test, if ( f(n) = \frac{1}{\ln(n)} ) is continuous, positive, and decreasing for ( n \geq 2 ), then the convergence of the integral ( \int_{2}^{\infty} f(x) , dx ) can be used to determine the behavior of the series.
- Set Up the Integral
We now find the integral: $$ \int_{2}^{\infty} \frac{1}{\ln(x)} , dx $$ To evaluate this integral, we can use a substitution.
- Substitution
Let ( u = \ln(x) ), thus ( du = \frac{1}{x} dx ), or ( dx = e^u , du ).
When ( x = 2 ), ( u = \ln(2) ). As ( x \to \infty ), ( u \to \infty ).
Now we rewrite the integral: $$ \int_{\ln(2)}^{\infty} \frac{1}{u} e^u , du $$
- Evaluate the Integral
The behavior of ( \int \frac{1}{u} , du ) is known to diverge as ( u \to \infty ). Therefore, since ( e^u ) grows faster than ( \frac{1}{u} ) can diminish, the integral diverges.
Thus, we conclude: Since the integral diverges, the series ( \sum_{n=2}^{\infty} \frac{1}{\ln(n)} ) also diverges by the Integral Test.
The series ( \sum_{n=2}^{\infty} \frac{1}{\ln(n)} ) diverges.
More Information
The divergence of the series ( \frac{1}{\ln(n)} ) highlights the behavior of logarithmic growth. As ( n ) increases, ( \ln(n) ) grows slowly, leading to the terms of the series diminishing slow enough that their sum does not converge.
Tips
- Misapplying the Integral Test: Be sure that the function is continuous, positive, and decreasing on the interval of integration.
- Confusing the growth rates of functions such as ( \ln(n) ) and polynomial or exponential functions, which can lead to incorrect conclusions about convergence.