Question

Use the root test to determine whether \(\sum_{m=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is as follows. $$ a_{n}=\frac{(\ln (1+\ln n))^{n}}{(\ln n)^{n}} $$

Short answer

The series \(\sum_{n=1}^{\infty} a_n\) converges.

Step-by-step solution

Identify the term to test

We are given the sequence \(a_n = \frac{(\ln(1 + \ln n))^n}{(\ln n)^n}\). Our task is to determine if the series \(\sum_{m=1}^{\infty} a_n\) converges by using the root test.

Apply the root test

The root test involves finding \(\lim_{n \to \infty} \sqrt[n]{|a_n|}\). For our sequence, \(|a_n| = \frac{(\ln(1 + \ln n))^n}{(\ln n)^n}\). By the root test, calculate this limit.

Simplify the expression

Since \(a_n\) is positive for sufficiently large \(n\), we have:\[\sqrt[n]{a_n} = \frac{\ln(1 + \ln n)}{\ln n}\]. Now, evaluate \(\lim_{n \to \infty} \frac{\ln(1 + \ln n)}{\ln n}\).

Analyze the limit

Consider the functions in the expression as \(n\) becomes very large:- \(\ln(1+\ln n)\) approximates to \(\ln(\ln n)\).- Therefore, the expression simplifies to \(\frac{\ln(\ln n)}{\ln n}\).Now, evaluate the limit: \[\lim_{n \to \infty} \frac{\ln(\ln n)}{\ln n} = 0\].

Apply the Root Test Conclusion

According to the root test, if the limit \(L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = 0\), the series \(\sum_{n=1}^{\infty} a_n\) converges absolutely.

Key concepts

Series Convergence

When we discuss "series convergence," we're essentially looking at whether the sum of an infinite series adds up to a finite number. For any series \(\sum_{n=1}^{\infty} a_n\), you can imagine adding all the terms \(a_1, a_2, a_3,... \) all the way to infinity. The magic happens when this infinite sum results in a finite value.
Convergence tells us that despite having an infinite number of terms, the series behaves well enough at infinity to settle on a finite sum.
- **Absolute convergence**: If the series \(\sum_{n=1}^{\infty} |a_n| \) (sum of the absolute values) converges, then the series \(\sum_{n=1}^{\infty} a_n \) is said to converge absolutely. Absolute convergence is a strong form of convergence.
- **Conditional convergence**: Occurs when the series \(\sum_{n=1}^{\infty} a_n \) converges, but \(\sum_{n=1}^{\infty} |a_n| \) does not.
In our exercise, we used the Root Test to determine convergence. If the limit found is less than 1, the series is absolutely converging.

Infinite Series

Infinite series consist of an unlimited number of terms added together. Mathematically, an infinite series is a sum of terms in the format \(S = a_1 + a_2 + a_3 + ...\). These terms are usually sequences, and one primary focus is whether or not this sum reaches a finite limit as the number of terms goes to infinity.
In our problem, the infinite series \(\sum_{m=1}^{\infty} a_n\) should be tested for its behavior as more terms are added. As \(n \to \infty\), the series will be finite only under specific mathematical conditions, such as the series being dominated by "small enough" terms.
Common tools to test series include:

• *Root Test:* Useful when dealing with terms raised to a power of \(n\). Here, calculating \(\lim_{n \to \infty} \sqrt[n]{|a_n|}\) demonstrates the series' behavior at infinity.
• *Ratio Test:* Another popular approach for exponential series and factorials.

By applying the Root Test, we determine if the terms of our series decrease fast enough to ensure the sum is finite, confirming its convergence.

Logarithmic Functions

Logarithmic functions are mathematical expressions that invert exponential functions, producing values that represent the power to which a fixed number, the base, must be raised to yield a given number. When we express them, they typically look like this: \( \ln(n)\), indicating the natural logarithm.
In the context of infinite series, logarithms are pivotal for handling power expressions precisely, as seen in our term \(a_n = \frac{(\ln(1+\ln n))^n}{(\ln n)^n}\). Here's a breakdown:

• The natural logarithm \( \ln(n)\) is used for calculating growth rates.
• In the given sequence, \(\ln(1+\ln n)\) modifies the growth rate, balancing out the series.

The manipulation of logarithmic functions can simplify series expressions, which is evident as \( n \) approaches infinity: simplifying involves approximations like using \( \ln(1+\ln n) \approx \ln(\ln n)\) for large \( n\).
Overall, understanding logarithmic functions helps unlock the complexity behind many geometric and exponential problems in series and calculus.