Question

Test the following series for convergence using the comparison test. (a) \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) Hint: Which is larger, \(n\) or \(\sqrt{n} ?\) (b) \(\sum_{n=2}^{\infty} \frac{1}{\ln n}\)

Short answer

Both series \[\frac{1}{\text{√n}}\] and \[\frac{1}{\text{ln} n}\] diverge.

Step-by-step solution

Identify the Series

The given series is \[\sum_{n=1}^{\text{∞}} \frac{1}{\text{√n}}\] and \[\sum_{n=2}^{\text{∞}} \frac{1}{\text{ln} n}\]. Each step will analyze the convergence of these series using the comparison test.

Analyze Series (a)

Consider \[\frac{1}{\text{√n}}\] and compare it to the harmonic series \[\frac{1}{n^{p}}\]. Here, \[\frac{1}{\text{√n}} = \frac{1}{n^{1/2}}\]. For the p-series test, if \[0 < p \text{ ≤} 1, \text{ the series diverges}\]. Since \[p = 1/2 \], the given series diverges.

Analyze Series (b)

Compare \[\frac{1}{\text{ln} n}\] with \[\frac{1}{n}\]. For large \[\text{n}\text{, ln n < n, thus } \frac{1}{\text{ln} n} > \frac{1}{n}\]. The harmonic series \[\frac{1}{n} \] diverges and \[\frac{1}{\text{ln} n}\] is of larger terms compared to \[\frac{1}{n} \] which means \[\text{it diverges}\].

Key concepts

divergence of series

In mathematical analysis, determining whether a series converges or diverges is a fundamental concept.
A series diverges if its terms do not approach zero or if the series approaches infinity as the number of terms increases.
This means the partial sums of the series do not converge to a finite limit.
For example, let's consider the series \(\frac{1}{\sqrt{n}}\).Since \(\frac{1}{\sqrt{n}}\) approaches zero, one might think it converges.
However, using the comparison test with another known divergent series, we see otherwise.
If we compare it to the series \(\frac{1}{n^{1/2}}\), which is a known p-series with p = 1/2, we see that p ≤ 1.
Therefore, the series diverges.
This logic also applies to more complex series, such as \(\frac{1}{\ln n}\).Even though the terms \(\frac{1}{\ln n}\) approach zero, we can compare it to the harmonic series \(\frac{1}{n}\).
Since \(\text{ln}(n) < n\), this means \(\frac{1}{\ln n} > \frac{1}{n}\), and we already know the series \(\frac{1}{n}\) diverges.
So, the series involving \(\frac{1}{\ln n}\) also diverges.
Understanding divergence is essential in series analysis and helps to quickly determine the behavior of a series.

p-series test

The p-series test is a handy method to determine the convergence or divergence of certain types of series.
A p-series is in the form \(\frac{1}{n^p}\), where p is a positive real number.
The test is based on the value of p:

• If \ p > 1 \, the series converges.
• If \ 0 < p \leq 1 , the series diverges.

Using this test, we can quickly analyze series like \(\frac{1}{n^{1/2}}\).Here, p = 1/2, which falls in the range where the series diverges.
This method is powerful because many series can be compared to a p-series to determine their behavior.
By understanding the effects of p, we gain insights into the series' convergence without needing to sum it explicitly.
Therefore, if we encounter a series that resembles \(\frac{1}{n^p}\), the p-series test gives a quick and accurate verdict on its convergence or divergence.

harmonic series

The harmonic series is one of the most well-known examples of a divergent series.
It is expressed as \(\frac{1}{n}\) and is called harmonic because it relates to harmonic functions in mathematical analysis.
Despite each term getting smaller and approaching zero, the harmonic series diverges.
The divergence can be surprising at first since the terms diminish, but their sum grows unbounded over time.
To be specific, the harmonic series diverges because the partial sums of its terms do not converge to a finite value.
This series serves as a benchmark for comparing other series, such as \(\frac{1}{\ln n}\).Since \(\frac{1}{\ln n}\) is greater than \(\frac{1}{n}\) for large n, it also diverges.
Understanding the harmonic series is crucial because it provides a clear example of how a series with diminishing terms can still diverge.
By recognizing patterns similar to the harmonic series, one can effectively use it in the comparison test to determine the behavior of more complex series.