Sequence Convergence: Does (11/n)^(11/n) Converge?

Hey guys! Let's dive into an interesting problem today: determining whether the sequence $a_n = (11/n)^{(11/n)}$ converges or diverges. If it converges, we'll also figure out its limit. This is a classic example in calculus where we need to use our knowledge of limits and sequences. So, grab your thinking caps, and let's get started!

Understanding Sequence Convergence and Divergence

Before we jump into the specifics of our sequence, let's quickly recap what it means for a sequence to converge or diverge. In the world of mathematics, understanding sequence convergence is crucial. A sequence {$a_n$} is said to converge if its terms get closer and closer to a specific value as n (the term number) approaches infinity. Mathematically, we write this as:

$$\lim_{n \to \infty} a_n = L$$

where L is the limit of the sequence. Basically, as n becomes incredibly large, $a_n$ gets arbitrarily close to L. Think of it like chasing a target; a convergent sequence is one that gets closer and closer to the bullseye.

On the flip side, a sequence diverges if its terms do not approach a specific value as n goes to infinity. This can happen in a few ways: the terms might oscillate wildly, grow without bound (approach infinity), or approach different values depending on how you look at the sequence. A divergent sequence is like a wild goose chase – there's no final destination.

To determine convergence or divergence, we often use limit techniques, and that's precisely what we'll do for our sequence $a_n = (11/n)^{(11/n)}$. Identifying the behavior of sequences as n approaches infinity is a fundamental concept in calculus and is essential for understanding more advanced topics like series and calculus. This concept allows mathematicians and scientists to model and predict the behavior of systems over time, making it a cornerstone of mathematical analysis.

Analyzing the Sequence $a_n = (11/n)^{(11/n)}$

Now, let's focus on the sequence $a_n = (11/n)^{(11/n)}$. To figure out if it converges or diverges, we need to evaluate the limit:

$$\lim_{n \to \infty} \left(\frac{11}{n}\right)^{\frac{11}{n}}$$

This looks a bit tricky, right? We have a variable in both the base and the exponent. A common strategy for dealing with such limits is to use logarithms. Logarithms are like mathematical detectives; they help us unravel complex expressions. Specifically, we'll take the natural logarithm (ln) of both the sequence and the limit.

Let's set:

$$y = \left(\frac{11}{n}\right)^{\frac{11}{n}}$$

Taking the natural logarithm of both sides, we get:

$$\ln(y) = \ln\left(\left(\frac{11}{n}\right)^{\frac{11}{n}}\right)$$

Using the logarithm power rule, which states that $\ln(a^b) = b \ln(a)$, we can simplify this to:

$$\ln(y) = \frac{11}{n} \ln\left(\frac{11}{n}\right)$$

Now we need to find the limit of $\ln(y)$ as n approaches infinity:

$$\lim_{n \to \infty} \ln(y) = \lim_{n \to \infty} \frac{11}{n} \ln\left(\frac{11}{n}\right)$$

This limit has the form 0 * (-∞), which is an indeterminate form. To handle this, we can rewrite the expression as a fraction so we can apply L'Hôpital's Rule. L'Hôpital's Rule is a powerful tool for finding limits of indeterminate forms. It allows us to take the derivatives of the numerator and denominator separately and then re-evaluate the limit.

Rewriting the expression, we get:

$$\lim_{n \to \infty} \ln(y) = 11 \lim_{n \to \infty} \frac{\ln\left(\frac{11}{n}\right)}{n}$$

Now we have the indeterminate form (-∞)/∞, perfect for L'Hôpital's Rule! Applying L'Hôpital's Rule means we differentiate the numerator and the denominator separately. This technique is incredibly useful when dealing with limits that initially result in indeterminate forms such as 0/0 or ∞/∞. It allows us to transform the limit into a more manageable form by focusing on the rates of change of the functions involved.

Applying L'Hôpital's Rule

Let's differentiate the numerator and the denominator separately:

• Numerator: The derivative of $\ln(11/n)$ with respect to n is:

  $$\frac{d}{dn} \ln\left(\frac{11}{n}\right) = \frac{d}{dn} (\ln(11) - \ln(n)) = 0 - \frac{1}{n} = -\frac{1}{n}$$

• Denominator: The derivative of n with respect to n is:

  $$\frac{d}{dn} n = 1$$

So, applying L'Hôpital's Rule, we get:

$$11 \lim_{n \to \infty} \frac{\ln\left(\frac{11}{n}\right)}{n} = 11 \lim_{n \to \infty} \frac{-\frac{1}{n}}{1} = 11 \lim_{n \to \infty} -\frac{1}{n}$$

Now this is much simpler! As n approaches infinity, -1/n approaches 0. Therefore:

$$\lim_{n \to \infty} \ln(y) = 11 \cdot 0 = 0$$

But remember, we found the limit of \ln(y), not y itself. To find the limit of y, we need to exponentiate both sides using the base e:

$$\lim_{n \to \infty} y = \lim_{n \to \infty} e^{\ln(y)} = e^{\lim_{n \to \infty} \ln(y)} = e^0 = 1$$

So, the limit of the sequence $a_n = (11/n)^{(11/n)}$ as n approaches infinity is 1.

Conclusion

Awesome! We've successfully determined that the sequence $a_n = (11/n)^{(11/n)}$ converges, and its limit is 1. We achieved this by using the power of logarithms to simplify the expression, dealing with the indeterminate form using L'Hôpital's Rule, and finally exponentiating to find the limit of the original sequence. This problem nicely illustrates how different mathematical tools can be combined to solve a seemingly complex problem.

Remember, guys, practice makes perfect! The more you work with limits and sequences, the more comfortable you'll become with these techniques. Keep exploring, keep learning, and most importantly, keep having fun with math! Understanding these fundamental concepts is crucial for advancing in calculus and other areas of mathematics. By mastering these techniques, you’re not just solving problems; you’re building a strong foundation for future mathematical endeavors.