Question

$\mathop {\lim }\limits_{n \to \infty } \left\{ {{{\left( {1 + \dfrac{1}{{{n^2}}}} \right)}^{\dfrac{2}{{{n^2}}}}}{{\left( {1 + \dfrac{{{2^2}}}{{{n^2}}}} \right)}^{\dfrac{4}{{{n^2}}}}}{{\left( {1 + \dfrac{{{3^2}}}{{{n^2}}}} \right)}^{\dfrac{6}{{{n^2}}}}}.....{{\left( {1 + \dfrac{{{n^2}}}{{{n^2}}}} \right)}^{\dfrac{{2n}}{{{n^2}}}}}} \right\} = $
A) $\dfrac{4}{e}$
B) $\dfrac{3}{e}$
C) $\dfrac{2}{e}$
D) $\dfrac{5}{e}$

Answer 1

Hint:
We can take the log of the given expression. Then we can simplify it using properties of logarithm and write it as a series. Then we can obtain a general term and write as a summation. Then we can find the sum of the limit by integration. After the integration, we can take the antilog to find the required limit.

Complete step by step solution:
We need to find the value of $\mathop {\lim }\limits_{n \to \infty } \left\{ {{{\left( {1 + \dfrac{1}{{{n^2}}}} \right)}^{\dfrac{2}{{{n^2}}}}}{{\left( {1 + \dfrac{{{2^2}}}{{{n^2}}}} \right)}^{\dfrac{4}{{{n^2}}}}}{{\left( {1 + \dfrac{{{3^2}}}{{{n^2}}}} \right)}^{\dfrac{6}{{{n^2}}}}}.....{{\left( {1 + \dfrac{{{n^2}}}{{{n^2}}}} \right)}^{\dfrac{{2n}}{{{n^2}}}}}} \right\}$
Let $I = \left\{ {{{\left( {1 + \dfrac{1}{{{n^2}}}} \right)}^{\dfrac{2}{{{n^2}}}}}{{\left( {1 + \dfrac{{{2^2}}}{{{n^2}}}} \right)}^{\dfrac{4}{{{n^2}}}}}{{\left( {1 + \dfrac{{{3^2}}}{{{n^2}}}} \right)}^{\dfrac{6}{{{n^2}}}}}.....{{\left( {1 + \dfrac{{{n^2}}}{{{n^2}}}} \right)}^{\dfrac{{2n}}{{{n^2}}}}}} \right\}$
We can take the logarithm of the term inside the limit.
 $ \Rightarrow \log I = \log \left\{ {{{\left( {1 + \dfrac{1}{{{n^2}}}} \right)}^{\dfrac{2}{{{n^2}}}}}{{\left( {1 + \dfrac{{{2^2}}}{{{n^2}}}} \right)}^{\dfrac{4}{{{n^2}}}}}{{\left( {1 + \dfrac{{{3^2}}}{{{n^2}}}} \right)}^{\dfrac{6}{{{n^2}}}}}.....{{\left( {1 + \dfrac{{{n^2}}}{{{n^2}}}} \right)}^{\dfrac{{2n}}{{{n^2}}}}}} \right\}$
We know that \[\log ab = \log a + \log b\] . So, we get,
 $ \Rightarrow \log I = \log {\left( {1 + \dfrac{1}{{{n^2}}}} \right)^{\dfrac{2}{{{n^2}}}}} + \log {\left( {1 + \dfrac{{{2^2}}}{{{n^2}}}} \right)^{\dfrac{4}{{{n^2}}}}} + \log {\left( {1 + \dfrac{{{3^2}}}{{{n^2}}}} \right)^{\dfrac{6}{{{n^2}}}}} + ..... + \log {\left( {1 + \dfrac{{{n^2}}}{{{n^2}}}} \right)^{\dfrac{{2n}}{{{n^2}}}}}$
Now we have a series with general term given by $\log {\left( {1 + \dfrac{{{r^2}}}{{{n^2}}}} \right)^{\dfrac{{2r}}{{{n^2}}}}}$ .
So, we can write its sum as,
 $ \Rightarrow \log I = \sum\limits_{r = 1}^n {\log {{\left( {1 + \dfrac{{{r^2}}}{{{n^2}}}} \right)}^{\dfrac{{2r}}{{{n^2}}}}}} $
We know that \[\log {a^b} = b\log a\] . So, the sum will become,
 $ \Rightarrow \log I = \sum\limits_{r = 1}^n {\dfrac{{2r}}{{{n^2}}}\log \left( {1 + \dfrac{{{r^2}}}{{{n^2}}}} \right)} $
Now we can take the limit
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\dfrac{{2r}}{{{n^2}}}\log \left( {1 + \dfrac{{{r^2}}}{{{n^2}}}} \right)} $
We know that the sum of limit tends to infinity by integrating from one to infinity.
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \int\limits_0^\infty {\dfrac{{2r}}{{{n^2}}}\log \left( {1 + \dfrac{{{r^2}}}{{{n^2}}}} \right)dr} $
We can give substitution as $x = \dfrac{r}{n}$ and on taking its derivative, \[\dfrac{{dx}}{{dr}} = \dfrac{1}{n} \Rightarrow \dfrac{{dr}}{n} = dx\]
As the limit is n tends to infinity, we can write limit of x from 0 to 1
On substituting, we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \int\limits_0^1 {2x\log \left( {1 + {x^2}} \right)dx} $
Now we can integrate by using the method of integrating by parts which is given by \[\int {u.v.dx} = {\text{ }}u\int {vdx} {\text{ }} - \int {u'\left( {\int {vdx} } \right)dx} \]
Here $u = \log \left( {1 + {x^2}} \right)$ and $v = 2x$
Then using chain rule, we get $u' = \dfrac{{2x}}{{1 + {x^2}}}$
And $\int {vdx} = 2\int {xdx} $
 $\int {vdx} = {x^2}$
On substituting these, the limit will become,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {{x^2}\log \left( {1 + {x^2}} \right)} \right]_0^1 - \int\limits_0^1 {2x \times \dfrac{{{x^2}}}{{1 + {x^2}}}dx} $
On applying the limit to the 1st part, we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {{1^2}\log \left( {1 + {1^2}} \right) - {0^2}\log \left( {1 + {0^2}} \right)} \right] - \int\limits_0^1 {\dfrac{{2{x^3}}}{{1 + {x^2}}}dx} $
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {\log 2} \right] - \int\limits_0^1 {\dfrac{{2{x^3}}}{{1 + {x^2}}}dx} $
We can expand ${x^3} = x\left( {1 + {x^2}} \right) - x$ . So, we get the limit as,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {\log 2} \right] - 2\int\limits_0^1 {\dfrac{{x\left( {1 + {x^2}} \right) - x}}{{1 + {x^2}}}dx} $
On splitting the numerator, we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {\log 2} \right] - 2\int\limits_0^1 {\dfrac{{x\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}dx} + 2\int\limits_0^1 {\dfrac{x}{{1 + {x^2}}}dx} $
On simplification, we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {\log 2} \right] - 2\int\limits_0^1 {xdx} + \int\limits_0^1 {\dfrac{{2x}}{{1 + {x^2}}}dx} $
We know that $\int\limits_{}^{} {xdx = \dfrac{{{x^2}}}{2}} $ and $\int\limits_{}^{} {\dfrac{{2x}}{{1 + {x^2}}}dx} = \log \left( {1 + {x^2}} \right)$
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {\log 2} \right] - \left[ {{x^2}} \right]_0^1 + \left[ {\log \left( {1 + {x^2}} \right)} \right]_0^1$
On applying the limits, we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {\log 2} \right] - \left[ {{1^2} - {0^2}} \right] + \left[ {\log \left( {1 + {1^2}} \right) - \log \left( {1 + 0} \right)} \right]$
We know that $\log 1 = 0$ , so we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \left[ {\log 2} \right] - 1 + \log \left( 2 \right)$
We know that $\log a + \log b = \log ab$ , so we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \log I = \log \left( 4 \right) - 1$
Now we have the limit of the log of the given expression. To find the required limit, we can take the antilog.
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } I = {e^{\log \left( 4 \right) - 1}}$
By properties of exponents, we can write,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } I = {e^{\log \left( 4 \right)}} \times {e^{ - 1}}$
We know that ${e^{\log a}} = a$ . So, we get,
 $ \Rightarrow \mathop {\lim }\limits_{n \to \infty } I = 4 \times \dfrac{1}{e}$
Thus, the required limit is $\dfrac{4}{e}$.

So, the correct answer is option A.

Note:
We cannot directly take the limit of the expression. We took the logarithm of the expression to express the product of n terms as the sum of their logarithms. We integrated the general term to find the summation. While giving substitution, we must make sure that we also change the limits of integration. We used the method of integration by parts for solving the integral. We must take the 2 functions in the correct order while applying integration by parts. We must take care of the signs of the terms after the integration.