Question

Use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison. $$\sum_{n=4}^{\infty} \frac{\ln n}{n-3}$$

Short answer

The series diverges. Compared with \( \sum \frac{1}{n} \ln n \), which also diverges.

Step-by-step solution

Identify the form of the given series

The given series is \( \sum_{n=4}^{\infty} \frac{\ln n}{n-3} \). This series takes a form where the numerator \( \ln n \) and the denominator \( n-3 \) suggest a comparison with a simpler series.

Choose a comparison series

For the Limit Comparison Test, we choose \( b_n = \frac{1}{n} \) because as \( n \to \infty \), \( \ln n \) is approximately \( n^0 \) and \( n-3 \) is approximately \( n \). The series \( \sum \frac{1}{n} \) is a harmonic series and is known to diverge.

Apply the Limit Comparison Test

Compute the following limit: \(\lim_{{n \to \infty}} \frac{a_n}{b_n} = \lim_{{n \to \infty}} \frac{\ln n}{n-3} \cdot n = \lim_{{n \to \infty}} \frac{n \ln n}{n-3} \).This limit simplifies to the form \(\lim_{{n \to \infty}} \frac{\ln n}{1-\frac{3}{n}} \approx \lim_{{n \to \infty}} \ln n \).Since \( \ln n \to \infty \), the limit is \( \infty \). This shows that simply dividing by a harmonic series is insufficient, suggesting reassessment.

Reassess and choose the correct comparison

Reconsider the choice of comparison series. A more appropriate choice should closely mirror the growth of \( \frac{\ln n}{n-3} \). We try \( b_n = \frac{1}{n} \ln n \) which behaves similarly for large \( n \).

Reapply the Limit Comparison Test

Calculate \( \lim_{{n \to \infty}} \frac{a_n}{b_n} = \lim_{{n \to \infty}} \frac{\ln n}{n-3} \cdot \frac{n}{\ln n} = \lim_{{n \to \infty}} \frac{n \ln n}{(n-3) \ln n} = 1 \).The limit is finite and non-zero, indicating the series \( \sum_{n=4}^{\infty} \frac{\ln n}{n-3} \) behaves like \( \sum_{n=4}^{\infty} \frac{1}{n} \ln n \), which diverges because it's a p-series with \( p = 1 \).

Conclusion

By the Limit Comparison Test, because the limit is a positive finite number, the original series \( \sum_{n=4}^{\infty} \frac{\ln n}{n-3} \) and \( \sum_{n=4}^{\infty} \frac{1}{n} \ln n \) have the same convergence behavior. Since \( \sum \frac{1}{n} \ln n \) diverges, the given series also diverges.

Key concepts

Convergence of Series

The convergence of a series is an important concept in calculus and analysis that determines whether the sum of an infinite sequence of numbers results in a finite number or diverges. A series \( \sum_{n=a}^{\infty} a_n \) is said to converge if, as we sum more and more terms, the total approaches a specific limit. Conversely, a series diverges if the sum does not settle towards any limit and continues to grow or oscillate indefinitely.

In this specific problem, we're tasked with determining the convergence of the series \( \sum_{n=4}^{\infty} \frac{\ln n}{n-3} \). Techniques such as the Limit Comparison Test help make these determinations by comparing the given series with another series of known behavior, thereby simplifying the convergence analysis.

Harmonic Series

The harmonic series is one of the most well-known examples of a divergent series in mathematics. It is defined as \( \sum_{n=1}^{\infty} \frac{1}{n} \). Despite the fact that the terms \( \frac{1}{n} \) approach zero, the harmonic series diverges because the sum of its infinite terms does not converge to any fixed number.

In our exercise, we initially considered \( \sum \frac{1}{n} \) to compare with the series \( \sum \frac{\ln n}{n-3} \). This choice helped highlight that the original form of the given series does not yield the expected comparison due to the nature of the logarithmic term. Although useful, the comparison showed that a more appropriate series for comparison was necessary for correct analysis.

P-Series

P-series are series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence or divergence of a p-series depends on the value of \( p \):

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

In our problem, the series \( \sum \frac{1}{n} \ln n \) behaves similarly to a p-series with \( p = 1 \) because adding the logarithmic function grows the denominator slower than \( n^{1+p} \). Thus \( \sum \frac{1}{n} \ln n \) diverges.

Using this understanding, we concluded that a direct comparison with \( \sum \frac{1}{n} \ln n \) accurately reflects the divergence of the given series since the limit was finite and non-zero.

ln Function

The natural logarithm function, denoted as \( \ln(n) \), is logarithm to the base \( e \), where \( e \approx 2.71828 \). This function describes how quantities grow multiplicatively. It's worth noting that the function grows very slowly and continues increasing as \( n \) becomes very large.

The \( \ln(n) \) in the numerator of our series \( \sum_{n=4}^{\infty} \frac{\ln n}{n-3} \) adds complexity to its convergence analysis. Unlike linear or power functions, the logarithmic function's gradual increase must be carefully considered when using tests like the Limit Comparison Test. Hence, initially a simple harmonic series failed to adequately match the behavior of this function. Ultimately, using \( \sum \frac{1}{n} \ln n \) provided a better basis for comparison because the logarithmic term matched components of the original function.