How To Solve Lim ⁡ N → ∞ ∏ K = 1 N Π Φ ( K ) ∏ K = 2 N ( Π K − 1 ) Φ ( K ) / K \lim_{n \to \infty} \frac{\prod_{k=1}^n \pi^{\varphi(k)}}{\prod_{k=2}^n \left( \pi^k - 1 \right)^{\varphi(k)/k}} Lim N → ∞ ​ ∏ K = 2 N ​ ( Π K − 1 ) Φ ( K ) / K ∏ K = 1 N ​ Π Φ ( K ) ​ ??

This article delves into the fascinating problem of evaluating the limit of a product involving the totient function. Specifically, we aim to solve the following limit:

$$\lim_{n \to \infty} \frac{\prod_{k=1}^n \pi^{\varphi(k)}}{\prod_{k=2}^n \left( \pi^k - 1 \right)^{\varphi(k)/k}}$$

where $\varphi$ denotes the Totient function (also known as Euler's totient function), and $\varphi(1) = 1$. This problem elegantly combines concepts from number theory, summation, and the properties of the totient function, making it a rich and rewarding mathematical exploration.

Understanding the Totient Function

Before diving into the solution, let's take a moment to understand the totient function, denoted by $\varphi(n)$. The totient function counts the number of positive integers less than or equal to $n$ that are relatively prime to $n$. In other words, $\varphi(n)$ gives the number of integers $k$ in the range $1 \le k \le n$ for which the greatest common divisor (GCD) of $n$ and $k$ is 1. The totient function is a cornerstone of number theory, with applications in cryptography, computer science, and various other fields.

For example, $\varphi(8) = 4$ because the numbers 1, 3, 5, and 7 are relatively prime to 8. Similarly, $\varphi(1) = 1$, $\varphi(2) = 1$, $\varphi(3) = 2$, $\varphi(4) = 2$, $\varphi(5) = 4$, and so on. A crucial property of the totient function is its multiplicativity: if $m$ and $n$ are relatively prime, then $\varphi(mn) = \varphi(m)\varphi(n)$. This property is instrumental in deriving formulas and simplifying expressions involving the totient function. Another essential identity is the formula for calculating $\varphi(n)$ based on the prime factorization of $n$. If $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ is the prime factorization of $n$, then

$$\varphi(n) = n \prod_{p|n} \left( 1 - \frac{1}{p} \right)$$

where the product is taken over all distinct prime numbers $p$ that divide $n$. This formula provides an efficient way to compute the totient function for any given integer $n$. Furthermore, the summation property of the totient function, $\sum_{d|n} \varphi(d) = n$, is particularly useful in many number-theoretic contexts. Understanding these fundamental properties of the totient function is crucial for tackling problems like the one presented in this article.

Transforming the Limit: The Logarithmic Approach

To tackle the given limit, a common and often effective strategy is to take the natural logarithm of the expression. This transforms the product into a sum, which is generally easier to manipulate. Let's denote the expression inside the limit as $L_n$:

$$L_n = \frac{\prod_{k=1}^n \pi^{\varphi(k)}}{\prod_{k=2}^n \left( \pi^k - 1 \right)^{\varphi(k)/k}}$$

Taking the natural logarithm of both sides, we get

$$\ln(L_n) = \ln \left( \frac{\prod_{k=1}^n \pi^{\varphi(k)}}{\prod_{k=2}^n \left( \pi^k - 1 \right)^{\varphi(k)/k}} \right)$$

Using the properties of logarithms, we can rewrite this as

$$\ln(L_n) = \sum_{k=1}^n \varphi(k) \ln(\pi) - \sum_{k=2}^n \frac{\varphi(k)}{k} \ln(\pi^k - 1)$$

Now, we can further manipulate the second term by using the logarithm property $\ln(ab) = \ln(a) + \ln(b)$ and the approximation $\ln(1-x) \approx -x$ for small $x$. We can rewrite $\ln(\pi^k - 1)$ as $\ln(\pi^k(1 - \pi^{-k})) = k\ln(\pi) + \ln(1 - \pi^{-k})$. Therefore,

$$\ln(L_n) = \sum_{k=1}^n \varphi(k) \ln(\pi) - \sum_{k=2}^n \frac{\varphi(k)}{k} \left( k\ln(\pi) + \ln(1 - \pi^{-k}) \right)$$

Expanding the second summation, we have

$$\ln(L_n) = \sum_{k=1}^n \varphi(k) \ln(\pi) - \sum_{k=2}^n \varphi(k) \ln(\pi) - \sum_{k=2}^n \frac{\varphi(k)}{k} \ln(1 - \pi^{-k})$$

The first two summations almost cancel out. Notice that the first sum includes the term for $k=1$, which is $\varphi(1)\ln(\pi) = \ln(\pi)$. Thus, we can rewrite the expression as

$$\ln(L_n) = \ln(\pi) - \sum_{k=2}^n \frac{\varphi(k)}{k} \ln(1 - \pi^{-k})$$

This transformation has significantly simplified the problem, converting the original limit of a product into a limit of a sum. Now, we can focus on analyzing the behavior of the remaining summation as $n$ approaches infinity.

Analyzing the Summation and Approximations

We now have the expression

$$\ln(L_n) = \ln(\pi) - \sum_{k=2}^n \frac{\varphi(k)}{k} \ln(1 - \pi^{-k})$$

To evaluate the limit as $n \to \infty$, we need to analyze the behavior of the summation term. Since $\pi > 1$, the term $\pi^{-k}$ approaches 0 as $k$ increases. This allows us to use the approximation $\ln(1 - x) \approx -x$ for small $x$. In our case, $x = \pi^{-k}$, so we have $\ln(1 - \pi^{-k}) \approx -\pi^{-k}$. Substituting this approximation into the summation, we get

$$\sum_{k=2}^n \frac{\varphi(k)}{k} \ln(1 - \pi^{-k}) \approx - \sum_{k=2}^n \frac{\varphi(k)}{k} \pi^{-k}$$

Thus, our expression for $\ln(L_n)$ becomes

$$\ln(L_n) \approx \ln(\pi) + \sum_{k=2}^n \frac{\varphi(k)}{k} \pi^{-k}$$

Now, we need to investigate the convergence of the series $\sum_{k=2}^{\infty} \frac{\varphi(k)}{k} \pi^{-k}$. To do this, we can use the fact that $\varphi(k) \le k$, which implies that $\frac{\varphi(k)}{k} \le 1$. Therefore,

$$\frac{\varphi(k)}{k} \pi^{-k} \le \pi^{-k}$$

The series $\sum_{k=2}^{\infty} \pi^{-k}$ is a geometric series with a common ratio of $\frac{1}{\pi}$, which is less than 1 since $\pi > 1$. Therefore, the geometric series converges. By the comparison test, the series $\sum_{k=2}^{\infty} \frac{\varphi(k)}{k} \pi^{-k}$ also converges. Let's denote the sum of this series as $S$:

$$S = \sum_{k=2}^{\infty} \frac{\varphi(k)}{k} \pi^{-k}$$

As $n$ approaches infinity, the summation $\sum_{k=2}^n \frac{\varphi(k)}{k} \pi^{-k}$ approaches $S$. Therefore, the limit of $\ln(L_n)$ as $n$ approaches infinity is

$$\lim_{n \to \infty} \ln(L_n) = \ln(\pi) + S$$

where $S$ is the convergent series $\sum_{k=2}^{\infty} \frac{\varphi(k)}{k} \pi^{-k}$.

Evaluating the Limit

To find the original limit, we need to exponentiate both sides of the equation:

$$\lim_{n \to \infty} L_n = \lim_{n \to \infty} e^{\ln(L_n)}$$

Since the exponential function is continuous, we can write

$$\lim_{n \to \infty} L_n = e^{\lim_{n \to \infty} \ln(L_n)}$$

Substituting the limit of $\ln(L_n)$ that we found earlier, we get

$$\lim_{n \to \infty} L_n = e^{\ln(\pi) + S} = e^{\ln(\pi)} e^S = \pi e^S$$

where $S = \sum_{k=2}^{\infty} \frac{\varphi(k)}{k} \pi^{-k}$. Thus, the final answer is

$$\lim_{n \to \infty} \frac{\prod_{k=1}^n \pi^{\varphi(k)}}{\prod_{k=2}^n \left( \pi^k - 1 \right)^{\varphi(k)/k}} = \pi e^{\sum_{k=2}^{\infty} \frac{\varphi(k)}{k} \pi^{-k}}$$

This result provides a closed-form expression for the limit, expressing it in terms of $\pi$ and an infinite series involving the totient function. The convergence of the series ensures that the limit is well-defined.

Conclusion

In this article, we successfully evaluated the limit of a product involving the totient function by employing a combination of logarithmic transformations, approximations, and series analysis. The key steps included taking the natural logarithm of the expression, simplifying the resulting summation using logarithmic properties, approximating $\ln(1 - x)$ with $-x$ for small $x$, and analyzing the convergence of the resulting series. The final result expresses the limit in terms of $\pi$ and an infinite series involving the totient function. This problem showcases the power and elegance of number theory, highlighting the importance of understanding fundamental concepts and employing appropriate techniques to solve complex mathematical problems. The process of converting the product into a sum via logarithms, and then leveraging approximations and series analysis, is a valuable strategy applicable to a wide range of mathematical problems. Furthermore, this exploration reinforces the significance of the totient function in number theory and its applications in various mathematical contexts. This problem not only provides a specific solution but also offers insights into general problem-solving techniques that are invaluable in mathematical research and education.