Test the series [ln(n)]^-1 for convergence.

In summary: For example, if you remove the first term from the harmonic series, you'll still get the harmonic series. What this means is that, although 10^9 seems like a lot, it's not really that big when you're talking about infinity.
  • #1
buffordboy23
548
2
Homework Statement

Test the series [tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} [/tex] for convergence.

The attempt at a solution

My thoughts are a comparison test with the harmonic series. It's clear that 1/ln(n) > 1/n for [tex] n \geq 2 [/tex]. This implies that

[tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=2} \frac{1}{n} = \left(\sum^{\infty}_{n=2} \frac{1}{n}\right) + 1 - 1 = \left(\sum^{\infty}_{n=1} \frac{1}{n}\right) - 1[/tex]Is it sufficient to say that

[tex] \left(\frac{1}{ln\left(2\right)} - \frac{1}{2}\right) + \left(\frac{1}{ln\left(3\right)} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{ln\left(N\right)} - \frac{1}{N}\right) \geq 1 [/tex]

for some integer N (I think N=3 or 4...have to check again), so that

[tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=1} \frac{1}{n} [/tex]

and so the series must diverge. This makes sense but I wonder if there is a simpler way to express the divergence.
 
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  • #2
buffordboy23 said:
Homework Statement

Test the series [tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} [/tex] for convergence.

The attempt at a solution

My thoughts are a comparison test with the harmonic series. It's clear that 1/ln(n) > 1/n for [tex] n \geq 2 [/tex]. This implies that

[tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=2} \frac{1}{n} = \left(\sum^{\infty}_{n=2} \frac{1}{n}\right) + 1 - 1 = \left(\sum^{\infty}_{n=1} \frac{1}{n}\right) - 1[/tex]
What you have above is all you need. The harmonic series diverges, and so does the same series with the first term removed. You have established that your series is term-be-term larger than the corresponding terms of the harmonic series (minus its first term), so your series diverges as well.
buffordboy23 said:
Is it sufficient to say that

[tex] \left(\frac{1}{ln\left(2\right)} - \frac{1}{2}\right) + \left(\frac{1}{ln\left(3\right)} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{ln\left(N\right)} - \frac{1}{N}\right) \geq 1 [/tex]

for some integer N (I think N=3 or 4...have to check again), so that

[tex]\sum^{\infty}_{n=2} \frac{1}{ln\left(n\right)} \geq \sum^{\infty}_{n=1} \frac{1}{n} [/tex]

and so the series must diverge. This makes sense but I wonder if there is a simpler way to express the divergence.
 
  • #3
Mark44 said:
What you have above is all you need. The harmonic series diverges, and so does the same series with the first term removed. You have established that your series is term-be-term larger than the corresponding terms of the harmonic series (minus its first term), so your series diverges as well.

Thanks for the feedback. What a trivial problem. I thought this was true, but it just didn't seem "strict" enough for proof.

So this suggests to me that the harmonic series is divergent for any starting index, say, n = 10^9, as n approaches infinity. This thought seems even more sketchy than the conventional n = 1 index on first appearance. I will work it out to see it, since it's been awhile since I did this stuff.
 
  • #4
So this suggests to me that the harmonic series is divergent for any starting index, say, n = 10^9, as n approaches infinity. This thought seems even more sketchy than the conventional n = 1 index on first appearance. I will work it out to see it, since it's been awhile since I did this stuff.
Yes, that's right. The thing is that although 10^9 seems pretty large, infinity is very much larger. The basic idea is that you can lop a finite number of terms from the beginning of a series without changing its basic behavior.
 

1. What does it mean to "test the series [ln(n)]^-1 for convergence?"

Testing the series [ln(n)]^-1 for convergence means determining if the series, which is an infinite sum of terms, will approach a finite limit or value as the number of terms goes to infinity. If it does, the series is said to be convergent. If it does not, the series is said to be divergent.

2. How do you test the series [ln(n)]^-1 for convergence?

To test the series [ln(n)]^-1 for convergence, you can use various convergence tests such as the integral test, comparison test, or ratio test. These tests involve manipulating the series and/or comparing it to known convergent or divergent series to determine the convergence or divergence of the given series.

3. What is the significance of the natural logarithm in [ln(n)]^-1?

The natural logarithm, denoted as ln(n), is the inverse of the exponential function e^x. It is a commonly used mathematical function in many scientific fields and has many applications in calculus and in the study of growth and decay processes. In the series [ln(n)]^-1, the natural logarithm is raised to the power of -1, which essentially means taking the reciprocal of ln(n).

4. What is the domain and range of [ln(n)]^-1?

The domain of [ln(n)]^-1 is all positive real numbers (i.e. n > 0) since the natural logarithm is only defined for positive numbers. The range of [ln(n)]^-1 is also all positive real numbers, as taking the reciprocal of a positive number will always result in a positive number.

5. What are some real-world applications of testing the series [ln(n)]^-1 for convergence?

Testing the series [ln(n)]^-1 for convergence has various applications in fields such as economics, physics, and engineering. For example, in economics, it can be used to analyze the rate of convergence of investments over time. In physics, it can be used to study the decay of radioactive materials. In engineering, it can be used to analyze the stability of a system over time.

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