limn→∞n−12(1+1n).(12.22.33....nn)1n2 is equal to

Let L=limn→∞n−12(1+1n).(11.22.33...nn)n−12In L=limn→∞−12(n+1n)ln n+1n2∑k=1nkln k =limn→∞−12(n+1n)ln n+1n2∑k=1n(k ln k−k ln n+k ln n) =limn→∞−12(n+1n)ln n+1n2∑k=1nk lnkn+ln nn2∑k=1nk =limn→∞−12(n+1n)ln n+1n∑k=1nkn lnkn+ln nn2.n(n+1)2

limn→∞n−12(1+1n).(12.22.33....nn)1n2 is equal to

Let L=limn→∞n−12(1+1n).(11.22.33...nn)n−12In L=limn→∞−12(n+1n)ln n+1n2∑k=1nkln k =limn→∞−12(n+1n)ln n+1n2∑k=1n(k ln k−k ln n+k ln n) =limn→∞−12(n+1n)ln n+1n2∑k=1nk lnkn+ln nn2∑k=1nk =limn→∞−12(n+1n)ln n+1n∑k=1nkn lnkn+ln nn2.n(n+1)2

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$\lim_{n \to \infty} n^{\frac{- 1}{2} (1 + \frac{1}{n})} . {(1^{2} {.2}^{2} {.3}^{3} .... n^{n})}^{\frac{1}{n^{2}}}$ is equal to

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$\frac{1}{\sqrt[4]{e}}$

$\sqrt{e}$

$\sqrt[4]{e}$

$\frac{1}{\sqrt{e}}$

answer is C.

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Detailed Solution

Let $L = \lim_{n \to \infty} n^{- \frac{1}{2} (1 + \frac{1}{n})} . {(1^{1} {.2}^{2} {.3}^{3} ... n^{n})}^{n^{- \frac{1}{2}}}$
In $L = \lim_{n \to \infty} - \frac{1}{2} (\frac{n + 1}{n}) \ln n + \frac{1}{n^{2}} \sum_{k = 1}^{n} k \ln k$
$= \lim_{n \to \infty} - \frac{1}{2} (\frac{n + 1}{n}) \ln n + \frac{1}{n^{2}} \sum_{k = 1}^{n} (k \ln k - k \ln n + k \ln n)$
$= \lim_{n \to \infty} - \frac{1}{2} (\frac{n + 1}{n}) \ln n + \frac{1}{n^{2}} \sum_{k = 1}^{n} k \ln \frac{k}{n} + \frac{\ln n}{n^{2}} \sum_{k = 1}^{n} k$
$= \lim_{n \to \infty} - \frac{1}{2} (\frac{n + 1}{n}) \ln n + \frac{1}{n} \sum_{k = 1}^{n} \frac{k}{n} \ln \frac{k}{n} + \frac{\ln n}{n^{2}} . \frac{n (n + 1)}{2}$