Question

The value of $\lim _{n \rightarrow \infty}\left(\frac{n !}{n^{n}}\right)^{\frac{2 n^{\prime}+1}{5 n^{2}+1}}$ is equal to (A) \(]\) (B) 0 (C) \(\left(\frac{1}{\mathrm{e}}\right)^{2 / 5}\) (D) \(\mathrm{e}^{2 / 3}\)

Short answer

Answer: (A) 1

Step-by-step solution

Rewrite the given expression using logarithms

Since we are dealing with exponentials, it would be easier to work with the logarithms and then convert back later. Let's rewrite the given expression in logarithmic form: Let L = \(\lim _{n \rightarrow \infty}\left(\frac{n!}{n^{n}}\right)^{\frac{2 n^{\prime}+1}{5 n^{2}+1}}\). Then \(\ln{L} = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \ln\left(\frac{n!}{n^n}\right)\)

Apply Stirling's Approximation

Stirling's approximation states that for large values of n, \(n! \approx n^ne^{-n}\sqrt{2\pi n}\). Now, let's substitute this approximation in the logarithmic expression: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \ln\left(\frac{n^ne^{-n}\sqrt{2\pi n}}{n^n}\right)\)

Simplify the expression

Now simplify the expression inside the logarithm and factor out constants: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \ln\left(e^{-n}\sqrt{2\pi n}\right) = \lim _{n \rightarrow \infty} \frac{2n+1}{5n^2+1} \left(-n + \frac{1}{2}\ln(2\pi n) \right)\)

Apply L'Hôpital's rule

Since we have the indeterminate form of the type \(\frac{\infty}{\infty}\), let's apply L'Hôpital's rule by taking derivatives of the numerator and denominator with respect to n: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{-1 + \frac{1}{2n}}{10n} = \lim _{n \rightarrow \infty} \frac{\frac{1}{2n} - 1}{10n}\) Now applying L'Hôpital's rule again: \(\ln{L} = \lim _{n \rightarrow \infty} \frac{-\frac{1}{2n^2}}{10} = 0\)

Calculate the final result

Since \(\ln{L} = 0\), we can find the value of L by taking the exponent with the base e: \(L = e^0 = 1\) So, the correct answer is: \(\lim _{n \rightarrow \infty}\left(\frac{n!}{n^{n}}\right)^{\frac{2 n^{\prime}+1}{5 n^{2}+1}} = 1\) which corresponds to the option (A).