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Evaluate the limit $$ y=\lim_{n\rightarrow \infty}n^{-\left(1+1/n\right)/2}\times \left(1^1\times 2^2\times 3^3\times\cdots\times n^n\right)^{1/{n^2}} $$

My Attempt:

When $n\rightarrow \infty$, then $n^{-\left(1+1/n\right)/2}\rightarrow n^{-1/2}$. So the limit becomes

$$ y=\lim_{n\rightarrow \infty}n^{-1/2}\times \left(1^1\times2^2\times3^3\times\cdots\times n^n\right)^{1/{n^2}} $$

Taking $\ln()$ on both side, we get

$$ \begin{align} y&=\lim_{n\rightarrow \infty}\ln\left\{n^{-1/2}\times\left(1^1\times2^2\times3^3\times\cdots\times n^n\right)^{1/{n^2}}\right\}\\ &=\lim_{n\rightarrow \infty}\left\{-\frac{1}{2}\ln(n)+\frac{1}{n^2}\ln\left(1^1\times2^2\times3^3\times\cdots\times n^n\right)\right\} \end{align} $$

How can I complete the solution from this point?

Fabrosi
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juantheron
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    Your line after "My Try" is wrong: if you pass to the limit $;n\to\infty;$ , the result cannot contain $;n;$ in it. Perhaps you tried some asymptotic approach? – Timbuc Oct 25 '14 at 02:45
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    See hyperfactorial $($ also here $)$. – Lucian Oct 25 '14 at 02:50
  • Maybe you can still use product of limits although your first limit is wrong. Its simple enough. The second limit in the product can be obtained by using the "sandwich" theorem. Just look at how the term $(1.2^2.3^2 \ldots n^n)^{\frac{1}{n^2}}$ is bounded. – Vishesh Oct 25 '14 at 02:56
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    One approach would be to rewrite your sequence as

    $$ \prod_{k=1}^{n} \left(\frac{k}{n}\right)^{k/n^2} = \left[ \prod_{k=1}^{n} \left(\frac{k}{n}\right)^{k/n} \right]^{1/n}, $$

    then take the log and consider the result as a Riemann sum. Specifically, it's equal to

    $$ \exp \left( \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \log \frac{k}{n} \right). $$

     – Antonio Vargas Oct 25 '14 at 02:58

2 Answers2

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As already commented by Lucian, $$\prod_{k=0}^n k^k=H(n)$$ which is the hyperfactorial function which also has a Stirling-like series $$H(n)=A e^{-\frac{n^2}{4}} n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}} \left(1+\frac{1}{720 n^2}-\frac{1433 }{7257600 n^4}+O\left(\left(\frac{1}{n}\right)^6\right)\right)$$ where $A$ is the Glaisher-Kinkelin constant.

Using it, the expression you are dealing with has then an asymptotic expansion $$\frac{1}{\sqrt[4]{e}}+\frac{12 \log (A)+\log (n)}{12 \sqrt[4]{e} n^2}+O\left(\left(\frac{1}{n}\right)^{7/2}\right)$$

Claude Leibovici
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In your last expression, the second term equals $$ \frac{1}{n^2} \sum_{k=1}^n k\ln k $$ (the term you dropped in your line "my try" does go to one, so this is correct). By using integrals to estimate the sum, we see that this equals $(1/2)\ln n -1/4 + o(1)$, so the original expression converges to $e^{-1/4}$.

To justify the evaluation of $\sum k\ln k$ as $\int_1^n x\ln x\, dx$, notice that the integrand has derivative $\lesssim \ln x$, so the mistake I'm making when I replace $k\ln k$ by the integral over $[k-1,k]$ is $\lesssim \ln k$ and thus the total error is $O(n\ln n)$.