Questions & Answers

What is the value of $\mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right).......3n}}{{{n^{2n}}}}} \right)^{\dfrac{1}{n}}}$ ?
(A) $\dfrac{{18}}{{{e^4}}}$
(B) $\dfrac{{27}}{{{e^2}}}$
(C) $\dfrac{9}{{{e^2}}}$
(D) $3\log 3 - 2$

Answer Verified Verified
Hint:Assume the limit as some variable and then try to separate the fraction into smaller fractions. Now take the natural log$\left( {{{\log }_e}x = \ln x} \right)$ of both sides. Use the properties of logarithmic function $\log {a^b} = b\log a$ and $\log abc = \log a + \log b + \log c$ carefully on the limit. Express the whole limit in form of summation. Now use proper substitution to express it in the form of integral.

Complete step-by-step answer:
Firstly, let’s put this whole limit equal to some $M$
$ \Rightarrow M = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right).......3n}}{{{n^{2n}}}}} \right)^{\dfrac{1}{n}}}$
Now we can rewrite the $3n$ in the numerator as $\left( {n + 2n} \right)$ and ${n^{2n}}$ in the denominator can also be expressed as $n \times n \times n........2n{\text{ times}}$. This will give us:
$ \Rightarrow M = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right).......3n}}{{{n^{2n}}}}} \right)^{\dfrac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{\left( {n + 1} \right)}}{n}\dfrac{{\left( {n + 2} \right)}}{n}......\dfrac{{\left( {n + 2n} \right)}}{n}} \right)^{\dfrac{1}{n}}}$
Since we see all the terms in multiplication, therefore logarithmic function will be the best option to use
$ \Rightarrow \ln M = \ln \left( {\mathop {\lim }\limits_{n \to \infty } {{\left( {\dfrac{{\left( {n + 1} \right)}}{n}\dfrac{{\left( {n + 2} \right)}}{n}......\dfrac{{\left( {n + 2n} \right)}}{n}} \right)}^{\dfrac{1}{n}}}} \right)$
By further simplifying this using the property of log, i.e. $\log {a^b} = b\log a$, we can write it as:
$ \Rightarrow \ln M = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\ln \left( {\dfrac{{\left( {n + 1} \right)}}{n}\dfrac{{\left( {n + 2} \right)}}{n}......\dfrac{{\left( {n + 2n} \right)}}{n}} \right)$
Also, we know the property of log in product to addition, i.e. $\log abc = \log a + \log b + \log c$. This can be of great use here:
$ \Rightarrow \ln M = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\left[ {\ln \dfrac{{\left( {n + 1} \right)}}{n} + \ln \dfrac{{\left( {n + 2} \right)}}{n}......... + \ln \dfrac{{\left( {n + 2n} \right)}}{n}} \right]$
So, we can notice a pattern in the sequence. This can be easily expression using summations for a finite number from $1$ to $2n$, and we will see:
$ \Rightarrow \ln M = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\sum\limits_{r = 1}^{2r} {\ln \dfrac{{\left( {n + r} \right)}}{n}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\sum\limits_{r = 1}^{2r} {\ln \left( {1 + \dfrac{r}{n}} \right)} $
Here, if we assume $\dfrac{r}{n} = x$ then we can say if $n \to \infty \Rightarrow x \to 0$ and also $\dfrac{r}{n} = x \Rightarrow \dfrac{1}{n} = dx$.
Then, for $r = 1 \Rightarrow x = 0$ and for $r = 2n \Rightarrow x = 2$
So, by using the above substitutions we can express this summation into integral with respect to the variable $x$ . We can write it as:
$ \Rightarrow \ln M = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{n}\sum\limits_{r = 1}^{2r} {\ln \left( {1 + \dfrac{r}{n}} \right)} = \int\limits_0^2 {\ln \left( {1 + x} \right)} dx$
Now this integral can be solved further to evaluate the value of M. We will use a method called integration by parts to solve the integration of $\ln \left( {1 + x} \right)$ . Integration by parts is a special method of integration that is often useful when two functions are multiplied together but is also helpful in other ways.
According to this method, $\int {uvdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}} \right)} } } \left( {\int {vdx} } \right)dx$ , where u and v are the functions in the variable $x$. Therefore,
$u=ln(1+x)$ and $v=1$ substituting in above formula
we get:
$ \Rightarrow \int {\ln \left( {1 + x} \right)dx = \ln \left( {1 + x} \right) \times\int 1 dx - \int {\dfrac{x}{{1 + x}}dx = \left[ {\ln \left( {1 + x} \right)x - x + \ln \left( {1 + x} \right)} \right]} } $

($\int \dfrac{x}{1 + x}dx$ can be solved by adding and subtracting 1 in the numerator and integrate it)
So, for our $\ln M$ , we can have:
$ \Rightarrow \ln M = \int\limits_0^2 {\ln \left( {1 + x} \right)} dx = \left[ {\ln \left( {1 + x} \right)x - x + \ln \left( {1 + x} \right)} \right]_0^2$
Now, let’s use the values of definite integral as $x = 0$ and $x = 2$
$ \Rightarrow \ln M = \left[ {\ln \left( {1 + x} \right)x - x + \ln \left( {1 + x} \right)} \right]_0^2 = 2\ln 3 - 2 + \ln 3 = \ln \left( {{3^2}} \right) - \ln \left( {{e^2}} \right) + \ln 3 = \ln \dfrac{{{3^2} \times 3}}{{{e^2}}}$
By the use of properties $\log {a^b} = b\log a$ and $\log abc = \log a + \log b + \log c$ again, we get to the expression:
 \[ \Rightarrow \ln M = \ln \dfrac{{{3^2} \times 3}}{{{e^2}}} = \ln \dfrac{{27}}{{{e^2}}}\]
Therefore, we can say: $M = \dfrac{{27}}{{{e^2}}}$
Hence, we get the solution to our limit $M = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right).......3n}}{{{n^{2n}}}}} \right)^{\dfrac{1}{n}}} = \dfrac{{27}}{{{e^2}}}$

So, the correct answer is “Option B”.

Note:Try to go step by step with the procedure to avoid any complications. Notice the part where the logarithmic function is taken in use. This function is very useful because of its property $\log {a^b} = b\log a$ and $\log abc = \log a + \log b + \log c$ , that can easily change any product into addition. Students should remember these formulas for solving these types of questions.