Limit

Challenge Problem:

Evaluate $\displaystyle \lim_{n \to \infty} \Bigg( n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $



Moderator Edit: Approved Challenge question.

Let me have a try


$\displaystyle
\lim_{n \to \infty} \Bigg( n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg)
$

$\displaystyle exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j\ln{j} \right] \}\
$

$\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j(\ln{j} - \ln{n}) + \frac{\ln{n}}{n^2} \sum_{j=1}^n j \right] \}\$

$\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{(n+1)\ln{n}}{2n}\right] + \int_0^1 x\ln{x}~dx \}\ $

$\displaystyle = exp \left( \int_0^1 x\ln{x}~dx \right) $

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

Quote:

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Originally Posted by simplependulum

Let me have a try


$\displaystyle
\lim_{n \to \infty} \Bigg( n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg)
$

$\displaystyle exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j\ln{j} \right] \}\
$

$\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j(\ln{j} - \ln{n}) + \frac{\ln{n}}{n^2} \sum_{j=1}^n j \right] \}\$

$\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{(n+1)\ln{n}}{2n}\right] + \int_0^1 x\ln{x}~dx \}\ $

$\displaystyle = exp \left( \int_0^1 x\ln{x}~dx \right) $

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

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That's basically my solution, too.


let $\displaystyle y = \ln \Bigg(n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $

then $\displaystyle \lim_{n \to \infty} y = \lim_{n \to \infty} \ln \Bigg(n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $

$\displaystyle = \lim_{n \to \infty} \Bigg( \ln n^{{-\frac{1}{2}(1+ \frac{1}{n})}} + \ln \Big( \prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $

$\displaystyle = \lim_{n \to \infty} \Big(-\frac{n+1}{2n} \ln n + \frac{1}{n^{2}} \sum_{j=1}^{n} \ln j^{j} \Big) $

$\displaystyle = \lim_{n \to \infty} \Big(- \frac{n(n+1)}{2n^{2}} \ln n + \frac{1}{n^{2}} \sum^{n}_{j=1} j \ln j \Big) $

$\displaystyle = \lim_{n \to \infty} \Big( - \frac{1}{n^{2}} \sum^{n}_{j=1} j \ln n + \frac{1}{n^{2}} \sum^{n}_{j=1} j \ln j \Big) $

$\displaystyle = \lim_{n \to \infty} \frac{1}{n^{2}} \sum_{j=1}^{n} (j \ln j - j \ln n) = \lim_{n \to \infty} \frac{1}{n^{2}} \sum_{j=1}^{n} j \ln \Big(\frac{j}{n} \Big) $

$\displaystyle = \lim_{n \to \infty} \frac{1}{n} \sum^{n}_{j=1} \frac{j}{n} \ln \Big(\frac{j}{n} \Big) = \lim_{b \to 0} \int^{1}_{b} x \ln x \ dx $

$\displaystyle = \lim_{b \to 0} \Big(\frac{x^{2}}{2} \ln x - \frac{x^{2}}{4}\Big) \Big|^{1}_{b} = - \frac{1}{4} $

so $\displaystyle \lim_{n \to \infty} \Bigg(n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg)= \lim_{n \to \infty} e^{y} = \frac{1}{\sqrt [4]{e}} $