Question

For each positive integer $n$ , let ${y_n} = \dfrac{1}{n}\left( {n + 1} \right)\left( {n + 2} \right)...{\left( {n + n} \right)^{\dfrac{1}{n}}}$ . For $x \in \mathbb{R}$ , let $\left[ x \right]$ be the greatest integer function less than or equal to $x$ , if $\mathop {\lim }\limits_{n \to \infty } {y_n} = L$ , then the value of $\left[ L \right]$ is _______ .

Answer 1

In this question, we are given a sequence ${y_n} = \dfrac{1}{n}\left( {n + 1} \right)\left( {n + 2} \right)...{\left( {n + n} \right)^{\dfrac{1}{n}}}$ , whose limit is $L$ at $n \to \infty$ .
First, we’ll take $n$ common and then, take ln on both sides. Then we will convert the summation into integration and finally integrate it to get the value of $L$ .
For a series in summation can be written as $\left( {a + 1} \right) + \left( {a + 2} \right) + ...\left( {a + n} \right) = \sum\limits_{r = 1}^n {\left( {a + r} \right)}$ .
For a given summation of the form $\mathop {\lim }\limits_{n \to \infty } \left[ {\dfrac{1}{n}\sum\limits_{r = 1}^n {\left( {1 + \dfrac{r}{n}} \right)} } \right]$ can be converted into integration as $\int\limits_0^1 {\left( {1 + x} \right)dx}$ by putting $\dfrac{r}{n} = x$ .
On differentiating with respect to $r$ , we get, $dx = \dfrac{1}{n}$ .
Lower limit $= \dfrac{{d\left( 1 \right)}}{{dn}} = 0$ and upper limit $= \dfrac{{d\left( n \right)}}{{dn}} = 1$ .

Complete answer:
Given series ${y_n} = \dfrac{1}{n}\left( {n + 1} \right)\left( {n + 2} \right)...{\left( {n + n} \right)^{\dfrac{1}{n}}}$ and it is also given that $\mathop {\lim }\limits_{n \to \infty } {y_n} = L$ .
To find the value of $\left[ L \right]$ .
First, taking $n$ common, we get.
$\begin{gathered} {y_n} = \dfrac{1}{n}{\left[ {{n^n}\left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{2}{n}} \right)...\left( {1 + \dfrac{n}{n}} \right)} \right]^{\dfrac{1}{n}}} \\\ {y_n} = \dfrac{n}{n}{\left[ {\left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{2}{n}} \right)...\left( {1 + \dfrac{n}{n}} \right)} \right]^{\dfrac{1}{n}}} \\\ {y_n} = {\left[ {\left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{2}{n}} \right)...\left( {1 + \dfrac{n}{n}} \right)} \right]^{\dfrac{1}{n}}} \\\ \end{gathered}$
Now, taking $\ln$ on both sides, we get,

\ln {y_n} = \ln {\left[ {\left( {1 + \dfrac{1}{n}} \right)\left( {1 + \dfrac{2}{n}} \right)...\left( {1 + \dfrac{n}{n}} \right)} \right]^{\dfrac{1}{n}}} \\\ \ln {y_n} = \dfrac{1}{n}\left[ {\ln \left( {1 + \dfrac{1}{n}} \right) + \ln \left( {1 + \dfrac{2}{n}} \right) + ...\ln \left( {1 + \dfrac{n}{n}} \right)} \right] \\\ \end{gathered} $$ which can be written as, $\ln {y_n} = \dfrac{1}{n}\sum\limits_{r = 1}^n {\ln \left( {1 + \dfrac{r}{n}} \right)} $ . Now, taking $\mathop {\lim }\limits_{n \to \infty } $ on both sides, we get, $$\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \mathop {\lim }\limits_{n \to \infty } \left[ {\dfrac{1}{n}\sum\limits_{r = 1}^n {\ln \left( {1 + \dfrac{r}{n}} \right)} } \right]$$ . Now, we will convert it into integration as explained above, we get, $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \int\limits_0^1 {\ln \left( {1 + x} \right)dx} $ . Now, integrating with respect to $x$ , we get, $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \left[ {x\ln (1 + x)} \right]_0^1 - \int\limits_0^1 {\dfrac{1}{{1 + x}} \cdot xdx} $ Adding and subtracting $1$ on the numerator, we get, $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \left[ {x\ln (1 + x)} \right]_0^1 - \int\limits_0^1 {\dfrac{{1 + x - 1}}{{1 + x}}dx} $ , which can be written as $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \left[ {x\ln (1 + x)} \right]_0^1 - \int\limits_0^1 {1 - \dfrac{1}{{1 + x}}dx} $ i.e., $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \left[ {x\ln (1 + x)} \right]_0^1 - \left[ {x - \ln (1 + x)} \right]_0^1$ . Finally, applying limits, we get, $\begin{gathered} \mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \left[ {1 \cdot \ln 2 - 0} \right] - \left[ {1 - \ln 2 - 0} \right] \\\ \mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \ln 2 - 1 + \ln 2 = 2\ln 2 - 1 \\\ \mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \ln 4 - 1 \\\ \end{gathered} $ Now, $1$ can be written as $\ln e$ , $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \ln 4 - \ln e$ . We know that, $\ln a - \ln b = \ln \dfrac{a}{b}$ , so using this property, we get, $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \ln \dfrac{4}{e}$ . Now, we have that, $\mathop {\lim }\limits_{n \to \infty } {y_n} = L$ , hence, replacing it, we get, $\mathop {\lim }\limits_{n \to \infty } \ln {y_n} = \ln \left( {\mathop {\lim }\limits_{n \to \infty } {y_n}} \right) = \ln L$ . Thus, $\ln L = \ln \dfrac{4}{e}$ , i.e., $L = \dfrac{4}{e}$ , whose values is approximately equal to $L = \dfrac{4}{{2.71}} = 1.47$ . Hence, the value for $\left[ L \right] = 1$ . **Note:** Properties of $\ln $ , which are $\ln {a^b} = b\ln a$ and $\ln a - \ln b = \ln \dfrac{a}{b}$ has to be kept in mind. Since, $\ln e = 1$ as the base for natural log is $e$ , so, we can replace $1$ by $\ln e$ . Value of $e = 2.718281828459045235602874713527$ , but while solving, we can round it to $100$ .