Question

$\lim _{n \rightarrow \infty}\left(\frac{n !}{(m n)^{n}}\right)^{1 / n}(m \in N)$ is equal to (A) \(1 / \mathrm{em}\) (B) \(\mathrm{m} / \mathrm{e}\) (C) em (D) \(\mathrm{e} / \mathrm{m}\)

Short answer

Short answer: Using Stirling's approximation and the properties of limits, we've found that the limit of \(\left(\frac{n!}{(mn)^{n}}\right)^{1/n}\) as n approaches infinity is \(\frac{1}{em}\), which corresponds to option (A).

Step-by-step solution

Rewrite the expression using natural logarithm and simplify

Take the natural logarithm of the sequence: \begin{align*} \ln\left(\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}}\right) &= \frac{1}{n}\ln\left(\frac{n!}{(mn)^n}\right). \end{align*}

Use Stirling's approximation

Stirling's approximation states that \(n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\) as n approaches infinity. Replace \(n!\) with the approximation: $$\frac{1}{n}\ln\left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{(mn)^n}\right).$$

Simplify the expression further

Simplify the expression inside the logarithm and then split the logarithm into the sum of logarithms. The simplified expression will be: \begin{align*} \frac{1}{n}\left[\ln\left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{(mn)^n}\right)\right] &= \frac{1}{n} \left[\frac{1}{2}\ln(2\pi n) + n\ln\left(\frac{n}{e}\right) - n\ln(mn)\right]. \end{align*}

Distribute 1/n inside the expression and find the limit

Distribute \(\frac{1}{n}\) inside the brackets and then find the limit as n approaches infinity. \begin{align*} \lim_{n\to\infty} \left[\frac{1}{2n}\ln(2\pi n) + \ln\left(\frac{n}{e}\right) - \ln(mn)\right] &= \lim_{n\to\infty} \left[\frac{1}{2n}\ln(2\pi n)\right] + \lim_{n\to\infty} \left[\ln\left(\frac{n}{e}\right)\right] - \lim_{n\to\infty} \left[\ln(mn)\right]. \end{align*} Now, notice that \(\lim_{n\to\infty} \left[\frac{1}{2n}\ln(2\pi n)\right] = 0\) since the logarithm grows slower than the linear term in the denominator. And, since m is a constant, we have \(\lim_{n\to\infty} \left[\ln(mn)\right] = \ln(m)\). So, the limit becomes: $$\lim_{n\to\infty} \left[\ln\left(\frac{n}{e}\right) - \ln(m)\right].$$

Combine the remaining limits into one logarithm

Combine the logarithms: $$\lim_{n\to\infty} \left[\ln\left(\frac{n}{me}\right)\right].$$

Take the exponential to get the original expression's limit

Take the exponential of both sides to reverse the process of taking the logarithm: $$\lim_{n\to\infty}\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty} e^{\left[\ln\left(\frac{n}{me}\right)\right]} = \frac{1}{m e}.$$ Hence, the correct answer is (A) \(\frac{1}{em}\).