Here is a report generated by Gemini Pro based on the content of this video. I found the video moved a little too quickly at times and the author read out every mathematical expression symbol by symbol which wasn't really necessary and it didn't aid comprehension. It's useful to be able to take the content of the video and put into a report that can be read at your leisure. If you want to tweak the LaTeX code, you can do so easily and I've done that below for a few of expressions.
The Limit of the nth Root: An Exploration of Euler's Sequence
At first glance, the limit of the sequence involving factorials and powers can seem intuitive, but as with many problems in analysis, the "obvious" answer can be misleading. This report dismantles the problem of finding the limit of the nth root of n factorial divided by n to the power of n. Following the path likely walked by Leonhard Euler, we move from discrete sequences to continuous calculus to find a beautiful truth connecting factorials, limits, and Euler’s number, e.
1. The Sequence and Intuition
We begin by defining our sequence \( a_n \):
$$ a_n = \frac{n!}{n^n} $$Intuitively, the denominator \( n^n \) grows much faster than the numerator \( n! \), suggesting that the terms drop toward zero. To prove this rigorously, we employ the Ratio Test. We investigate the ratio of consecutive terms:
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} $$Simplifying the factorials (\( (n+1)! = (n+1)n! \)) and regrouping terms gives:
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left( \frac{n}{n+1} \right)^n $$This expression can be rewritten to reveal a fundamental definition of Euler's number:
$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{\left( \frac{n+1}{n} \right)^n} = \lim_{n \to \infty} \frac{1}{\left( 1 + \frac{1}{n} \right)^n} = \frac{1}{e} $$Since \( \dfrac{1}{e} \approx 0.3679 \) is strictly less than 1, the Ratio Test confirms that the sequence converges to 0.
2. The Main Problem: The nth Root
Establishing that the base sequence goes to 0 is just the preamble. The true challenge is finding the limit of the nth root of this sequence:
$$ L = \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} $$Since the base approaches 0 and the exponent \( \frac{1}{n} \) also approaches 0, we are facing the indeterminate form \( 0^0 \). To resolve this, we use the natural logarithm to transform the product into a sum, bridging the gap between discrete algebra and integral calculus.
3. From Logarithms to Riemann Sums
Let \( y = \sqrt[ \uproot{4} n]{a_n} \). Taking the natural log of the limit allows us to work with sums:
$$ \ln(L) = \lim_{n \to \infty} \ln \left( \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} \right) = \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) $$Using logarithm rules, we expand the term inside:
$$ \ln \left( \frac{n!}{n^n} \right) = \ln(n!) - \ln(n^n) = \sum_{k=1}^{n} \ln(k) - n \ln(n) $$We can cleverly rewrite \( n \ln(n) \) as a sum of \( n \) identical terms: \( \displaystyle \sum_{k=1}^{n} \ln(n) \). Substituting this back into our limit equation gives:
$$ \ln(L) = \lim_{n \to \infty} \frac{1}{n} \left( \sum_{k=1}^{n} \ln(k) - \sum_{k=1}^{n} \ln(n) \right) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( \ln(k) - \ln(n) \right) $$Combining the logs, we arrive at a recognizable form:
$$ \ln(L) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} \right) $$This structure—a sum of function values multiplied by a small width \( \frac{1}{n} \) is precisely the definition of a Riemann Sum on the interval \([0, 1]\). As \( n \to \infty \), the sum converges to a definite integral:
4. Evaluating the Integral
We now evaluate the improper integral of \( \ln(x) \) from 0 to 1. Using integration by parts (where \( u = \ln(x) \) and \( dv = dx \)), we get:
$$ \int \ln(x) \, dx = x \ln(x) - x $$We evaluate the definite integral by taking the limit as the lower bound approaches 0:
$$ \Bigl[ x \ln(x) - x \Bigr]_0^1 = (1 \ln(1) - 1) - \lim_{a \to 0^+} (a \ln(a) - a) $$We know \( \ln(1) = 0 \), so the first term is \( -1 \). For the limit \( \lim_{a \to 0^+} a \ln(a) \), we use L'Hôpital's rule, which shows that this term vanishes to 0. Thus:
$$ \ln(L) = -1 - 0 = -1 $$Conclusion
We have found that the natural logarithm of the limit is \( -1 \). To find the original limit \( L \), we simply exponentiate both sides:
$$ L = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}}=e^{-1} = \frac{1}{e} $$We have proven that while the sequence \( \dfrac{n!}{n^n} \) collapses to zero, its geometric mean—the nth root—stabilizes at exactly 1 over e. This problem serves as an elegant bridge connecting basic sequence properties, the definition of e, and the power of integral calculus.
