Could you please tell me how it was proved that the limit of e^ (n)*n!/n^ (n) is infinity?

The limit of the sequence \( a_n = \frac{e^n n!}{n^n} \) as \( n \to \infty \) is proven to be infinity through a rigorous application of Stirling's approximation for the factorial function, combined with careful asymptotic analysis. Stirling's approximation provides the foundational estimate \( n! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \), meaning the ratio of these two expressions tends to 1 as \( n \) grows. Substituting this leading-order approximation directly into the expression for \( a_n \) yields \( a_n \sim \frac{e^n \cdot \sqrt{2\pi n} (n/e)^n}{n^n} = \sqrt{2\pi n} \). This simplified form clearly diverges to infinity, as the square root term grows without bound. However, this direct substitution only shows that a sequence asymptotically equivalent to \( a_n \) diverges; a complete proof must demonstrate that the error inherent in Stirling's approximation does not alter this divergent conclusion.

A more meticulous proof, often employing the logarithm of the sequence and integral comparisons, solidifies this result. By taking the natural log, we examine \( \ln(a_n) = n + \ln(n!) - n\ln(n) \). The term \( \ln(n!) \) can be bounded using the integral \( \int_1^n \ln(x) \, dx = n\ln(n) - n + 1 \). The well-known inequality \( \int_1^n \ln(x) \, dx \le \ln(n!) \le \int_1^{n+1} \ln(x) \, dx \) provides concrete bounds. Applying these to the expression for \( \ln(a_n) \) shows it is bounded below by a term like \( 1 - \ln(n) \) and above by \( \ln(1 + 1/n)^n + \ln(n+1) \). While the lower bound tends to negative infinity, it is too weak; a sharper analysis using the Euler-Maclaurin expansion or a more precise inequality reveals the dominant growth term. Crucially, the more precise form of Stirling's series, \( n! = \sqrt{2\pi n} (n/e)^n e^{\mu_n} \) where \( \mu_n \to 0 \), can be inserted directly. This gives \( a_n = \sqrt{2\pi n} \, e^{\mu_n} \), which explicitly demonstrates divergence because \( e^{\mu_n} \) tends to a finite positive constant.

The implication of this divergence is significant in probability, statistics, and analysis, as this sequence appears in Stirling's formula itself and in evaluations of certain series. Its growth, proportional to \( \sqrt{n} \), indicates that \( n! \) ultimately grows faster than \( (n/e)^n \) by a multiplicative factor of \( e^n \sqrt{2\pi n} / n^n \) in this specific arrangement. This result is counterintuitive to some because the numerator and denominator both contain super-exponential components; the proof clarifies that the numerator's \( e^n n! \) term dominates the denominator's \( n^n \) term precisely due to the hidden \( \sqrt{n} \) factor emerging from the factorial's deep structure. Consequently, any asymptotic comparison of factorial growth to exponential forms must account for this polynomial-order correction factor.

Therefore, the limit is definitively infinity, proven by establishing that \( a_n \) is asymptotically equivalent to, or bounded below by, a sequence that scales with \( \sqrt{n} \). The most efficient and conclusive method remains the direct application of Stirling's approximation in its full form, \( n! = \sqrt{2\pi n} (n/e)^n (1 + O(1/n)) \), which immediately yields \( a_n = \sqrt{2\pi n} (1 + O(1/n)) \), a sequence whose limit behavior is unequivocal. This approach avoids the pitfalls of crude bounding and leverages a established, verifiable asymptotic expansion to provide a clean, authoritative resolution.