How can I evalute this product?

Question

How can I evalute this product??

$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}}$$

Unfortunately, I have no idea.

Answer 1

Hint:

Taking the logarithm,

$$\log p(n)=-\log n\sum_{i=1}^\infty i\,n^{-i}.$$

This summation, which is a modified geometric series, has a closed-form formula.

Answer 2

\begin{eqnarray*} P=\prod_{i=1}^{\infty} (n^{-i})^{n^{-i}} = \prod_{i=1}^{\infty} n^{-in^{-i}} = n^{-\sum_{i=1}^{\infty}in^{-i}} \end{eqnarray*} $\sum_{i=1}^{\infty}ix^{i}= \frac{x}{(1-x)^2}$ \begin{eqnarray*} P=n^{-\frac{n}{(n-1)^2}} \end{eqnarray*}

Answer 3

HINT:

$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}} = \prod_{i=1}^{\infty} \frac{1}{n^{\frac{i}{n^i}}} = \frac{1}{n^{\frac{1}{n}}} \cdot \frac{1}{n^{\frac{2}{n^2}}}\cdot\frac{1}{n^{\frac{3}{n^3}}}...$$

$$\prod_{i=1}^{k} {(n^{-i})}^{n^{-i}} = \frac{1}{n^{\frac{n^{k-1} + 2 \cdot n^{k-2}...(k-1) \cdot n + k}{n^k}}}$$

And the exponent can be made into a summation.

Answer 4

Taking log $$\ln\Big( \prod_{i=1}^{\infty} (n^{-i})^{n^{-i}}\Big)$$$$$$ $$\sum_{i=1}^{\infty} \frac{i}{n^i} \ln n$$$$$$ $$\ln n \sum_{i=1}^{\infty} \frac{i}{n^i} $$$$$$

Let$$$$$$S= \sum_{i=1}^{k} \frac{i}{n^i}$$ $$$$ $$S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\cdots+\frac{k}{n^k}$$$$$$ Multiplying by $\frac{1}{n}$$$$$ $$\frac{S}{n}=\frac{1}{n^2}+\frac{2}{n^3}+\frac{3}{n^4}+\cdots+\frac{k}{n^{k+1}}$$$$$$ Now $S-\frac{S}{n}$$$$$ $$\frac{n-1}{n} S =\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\cdots+\frac{1}{n^k}-\frac{k}{n^{k+1}}$$$$$$ Using GP formula, $$\frac{n-1}{n} S =\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}$$$$$$ $$S=\frac{n}{n-1} \Biggr(\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}\Biggr)$$$$$$ For $k=\infty$ if $n\gt 1$$$$$ $$S=\frac{n}{n-1}\frac{1}{n} \Big(\frac{n}{n-1}\Big)$$$$$$ $$S=n\Big(\frac{1}{n-1}\Big)^2$$$$$$

Then substitute it back

$$n \ln n \Big(\frac{1}{n-1}\Big)^2$$Taking antilog,$$n^{n\Big(\frac{1}{n-1}\Big)^2}$$

This is only for $n\gt 1$