Question

Consider the following \displaystyle \lim_{n \to \infty }{{\left\\{ \dfrac{n!}{{{\left( kn \right)}^{n}}} \right\\}}^{\dfrac{1}{n}}}, $k\ne 0$, what is it equal to?
(a) $\dfrac{k}{e}$
(b) $\dfrac{e}{k}$
(c) $\dfrac{1}{ke}$
(d) None of these

Answer 1

Use the formula $n!=n\left( n-1 \right)\left( n-2 \right)......1$ and take n common from all the terms and write it in the exponential form. Cancel the common term ${{n}^{n}}$ from the numerator and the denominator. Now, take natural log both the sides and use the formula $\log {{a}^{m}}=m\log a$ to simplify. Also, apply the formula $\log \left( \dfrac{m}{n} \right)=\log m-\log n$ to break the terms. Convert the given sum of infinite terms into a summation series of the form $\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n-1}{\dfrac{1}{n}f\left( \dfrac{r}{n} \right)}$. Now, in the next step convert this expression of limit into a definite integral by replacing $\left( \dfrac{r}{n} \right)$ with x and $\dfrac{1}{n}$ with $dx$. The upper and lower limits of the integral will be found by substituting r = n – 1 and r = 1 in the expression $\displaystyle \lim_{n \to \infty }\left( \dfrac{r}{n} \right)$ respectively and simplifying. Finally, use the formula $\int{\log \left( ax+b \right)dx}=\dfrac{1}{a}\left( ax+b \right)\left( \log \left( ax+b \right)-1 \right)$ to evaluate the integral and substitute the limits to get the answer.
Here we have been asked to find the value of the expression \displaystyle \lim_{n \to \infty }{{\left\\{ \dfrac{n!}{{{\left( kn \right)}^{n}}} \right\\}}^{\dfrac{1}{n}}} where $k\ne 0$.

Complete step by step answer:
Now, we need to convert the expression of limit into a definite integral to solve the question. Let us assume the expression as $I$.
\Rightarrow I=\displaystyle \lim_{n \to \infty }{{\left\\{ \dfrac{n!}{{{\left( kn \right)}^{n}}} \right\\}}^{\dfrac{1}{n}}}
Using the formula $n!=n\left( n-1 \right)\left( n-2 \right)......1$ we get,
\Rightarrow I=\displaystyle \lim_{n \to \infty }{{\left\\{ \dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)......1}{{{\left( kn \right)}^{n}}} \right\\}}^{\dfrac{1}{n}}}
Since there are n terms in the numerator of the limit so taking n common from each term we get,
\Rightarrow I=\displaystyle \lim_{n \to \infty }{{\left\\{ \dfrac{{{n}^{n}}\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\dfrac{1}{n}}{{{\left( kn \right)}^{n}}} \right\\}}^{\dfrac{1}{n}}}
Cancelling the common factor and simplifying the numerator in a more specific pattern we get,
\Rightarrow I=\displaystyle \lim_{n \to \infty }{{\left\\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)}{{{k}^{n}}} \right\\}}^{\dfrac{1}{n}}}
Taking natural log, i.e. log to the base e on both the sides and using the formula $\log {{a}^{m}}=m\log a$ we get,
\begin{aligned} & \Rightarrow \log I=\log \left[ \displaystyle \lim_{n \to \infty }{{\left\\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)}{{{k}^{n}}} \right\\}}^{\dfrac{1}{n}}} \right] \\\ & \Rightarrow \log I=\displaystyle \lim_{n \to \infty }\left[ \log {{\left\\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)}{{{k}^{n}}} \right\\}}^{\dfrac{1}{n}}} \right] \\\ & \Rightarrow \log I=\displaystyle \lim_{n \to \infty }\left[ \dfrac{1}{n}\log \left\\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)}{{{k}^{n}}} \right\\} \right] \\\ \end{aligned}
Further using the formula $\log \left( \dfrac{m}{n} \right)=\log m-\log n$ we can break the terms of logarithm, so we get,

& \Rightarrow \log I=\displaystyle \lim_{n \to \infty }\left[ \dfrac{1}{n}\left\\{ \log \left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)-\log {{k}^{n}} \right\\} \right] \\\ & \Rightarrow \log I=\displaystyle \lim_{n \to \infty }\left[ \dfrac{1}{n}\left\\{ \log \left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)-n\log k \right\\} \right] \\\ & \Rightarrow \log I=\displaystyle \lim_{n \to \infty }\left[ \dfrac{1}{n}\log \left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)-\log k \right] \\\ \end{aligned}$$ Multiplying the limit with each term by removing the bracket we get, $$\Rightarrow \log I=\displaystyle \lim_{n \to \infty }\dfrac{1}{n}\log \left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)-\displaystyle \lim_{n \to \infty }\left( \log k \right)$$ Since $\log k$ is a constant and it does not contain the terms of n so the limit $\displaystyle \lim_{n \to \infty }$ will have no effect on it and therefore we can write the above expression as: $$\begin{aligned} & \Rightarrow \log I=\displaystyle \lim_{n \to \infty }\dfrac{1}{n}\log \left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)-\log k \\\ & \Rightarrow \log I+\log k=\displaystyle \lim_{n \to \infty }\dfrac{1}{n}\log \left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right) \\\ \end{aligned}$$ Using the formula $\log m+\log n=\log \left( mn \right)$ in the L.H.S we get, $$\Rightarrow \log \left( kI \right)=\displaystyle \lim_{n \to \infty }\dfrac{1}{n}\log \left( 1-\dfrac{1}{n} \right)\left( 1-\dfrac{2}{n} \right)\left( 1-\dfrac{3}{n} \right)......\left( 1-\dfrac{\left( n-1 \right)}{n} \right)$$ We can write the above expression in the summation form as: $$\Rightarrow \log \left( kI \right)=\displaystyle \lim_{n \to \infty }\dfrac{1}{n}\sum\limits_{r=1}^{n-1}{\log \left( 1-\dfrac{r}{n} \right)}$$ Now, we know that any limit of the form $\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n-1}{\dfrac{1}{n}f\left( \dfrac{r}{n} \right)}$ can be converted into a definite integral by replacing $\left( \dfrac{r}{n} \right)$ with x and $\dfrac{1}{n}$ with $dx$. So, the limit can be written as: $$\Rightarrow \log \left( kI \right)=\int{\log \left( 1-x \right)dx}$$ Here, we need to determine the limits of this integral. To do this we substitute r = n – 1 and r = 1 in the expression $\displaystyle \lim_{n \to \infty }\left( \dfrac{r}{n} \right)$ respectively and simplify. So the upper limit will be $\displaystyle \lim_{n \to \infty }\left( \dfrac{n-1}{n} \right)=\displaystyle \lim_{n \to \infty }\left( 1-\dfrac{1}{n} \right)=1$ and the lower limit will be $\displaystyle \lim_{n \to \infty }\left( \dfrac{1}{n} \right)=0$. Therefore the integral becomes: $$\Rightarrow \log \left( kI \right)=\int_{0}^{1}{\log \left( 1-x \right)dx}$$ Using the formula $$\int{\log \left( ax+b \right)dx}=\dfrac{1}{a}\left( ax+b \right)\left( \log \left( ax+b \right)-1 \right)$$ where $$a=-1$$ and $b=1$ we get, $$\begin{aligned} & \Rightarrow \log \left( kI \right)=\left( \dfrac{1}{-1} \right)\left[ \left( 1-x \right)\left( \log \left( 1-x \right)-1 \right) \right]_{0}^{1} \\\ & \Rightarrow \log \left( kI \right)=-1\left[ \left( 1-x \right)\left( \log \left( 1-x \right)-1 \right) \right]_{0}^{1} \\\ \end{aligned}$$ Substituting the values of limits we get, $$\begin{aligned} & \Rightarrow \log \left( kI \right)=-1\left[ \left( 1-1 \right)\left( \log \left( 1-1 \right)-1 \right)-\left( 1-0 \right)\left( \log \left( 1-0 \right)-1 \right) \right] \\\ & \Rightarrow \log \left( kI \right)=-1\left[ \left( 0 \right)\left( \log \left( 0 \right)-1 \right)-\left( 1 \right)\left( \log \left( 1 \right)-1 \right) \right] \\\ \end{aligned}$$ We know that $\log 1=0$, so we get, $$\begin{aligned} & \Rightarrow \log \left( kI \right)=-1\left[ -\left( 1 \right)\left( -1 \right) \right] \\\ & \Rightarrow \log \left( kI \right)=-1 \\\ \end{aligned}$$ We can write $-1=\log \left( \dfrac{1}{e} \right)$ so using this relation in the R.H.S we get, $$\begin{aligned} & \Rightarrow \log \left( kI \right)=-1\left[ -\left( 1 \right)\left( -1 \right) \right] \\\ & \Rightarrow \log \left( kI \right)=\log \left( \dfrac{1}{e} \right) \\\ \end{aligned}$$ Removing the log function from both the sides we get, $$\begin{aligned} & \Rightarrow kI=\dfrac{1}{e} \\\ & \therefore I=\dfrac{1}{ke} \\\ \end{aligned}$$ **So, the correct answer is “Option c”.** **Note:** Note that all the base of log function is e everywhere in the above solution so do not consider it as 10. We cannot directly substitute $n=\infty $ in the expression to get the answer because initially it is in indeterminate form. You may get the idea when to convert the limit into definite integral by observing that the exponent of the expression as a whole is $\dfrac{1}{n}$ that means it can be replaced with x once we are able to remove it from the exponent by taking log.