Equivalent of a product

Is there an asymptotic equivalent for $$\prod_{i=1}^k i^i$$ ?

Can we find something better than $k^{k^2}$ ?

Thank you

Solution 1:

It is possible to compute equivalent of $\displaystyle{P_n=\prod_{k=1}^nk^k}$ by elementary means. Here is how :

First compute $S_n=\ln(P_n)$ :

$$S_n=\sum_{k=2}^nk\ln(k)$$

Using the fact that $x\mapsto x\ln(x)$ is increasing on $[\exp(-1),+\infty)$, we get a first (rough) estimate :

$$\int_1^nt\ln(t)\,dt\le S_n\le\int_1^{n+1}t\ln(t)\,dt$$

which leads, after a few computations, to :

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+o(n^2)$$

Now the goal is to refine this expansion until precision $o(1)$ is reached, so that we shall take the exponential and get an equivalent for $P_n$.

Let $$x_n=S_n-\frac{n^2}2\ln(n)+\frac{n^2}4$$

We have

$$x_n-x_{n-1}=n\ln(n)-\frac{n^2}2\ln(n)+\frac{n^2}4+\frac{(n-1)^2}2\ln(n-1)-\frac{(n-1)^2}4\sim\frac12\ln(n)$$

Hence, by partial summation of positive divergent series : $x_n\sim \frac n2\ln(n)$, so that

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+\frac n2\ln(n)+o(n\ln(n))$$

Next let :

$$y_n=S_n-\frac{n^2}2\ln(n)+\frac{n^2}4-\frac n2\ln(n)$$

We have :

$$y_n-y_{n-1}\sim\frac1{12n}$$

Hence, again by partial summation of positive divergent series :

$$y_n\sim\frac{1}{12}\ln(n)$$

so that :

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+\frac n2\ln(n)+\frac{1}{12}\ln(n)+o(\ln(n))$$

Next, let :

$$z_n=S_n-\frac{n^2}2\ln(n)+\frac{n^2}4-\frac n2\ln(n)-\frac{1}{12}\ln(n)$$

We have :

$$z_n-z_{n+1}\sim\frac1{360n^3}$$

which proves the convergence of the series $\sum z_n$ and hence the convergence of the sequence $(z_n)$ to some limit $L\in\mathbb{R}$.

So we get :

$$S_n=\frac{n^2}2\ln(n)-\frac{n^2}4+\frac n2\ln(n)+\frac{1}{12}\ln(n)+L+o(1)$$

It's time to come back to $P_n$ :

$$\boxed{P_n=\lambda e^{-n^2/4}n^{\frac{n(n+1)}{2}+\frac1{12}}}$$

where $\lambda=e^L$

Solution 2:

These numbers are tabulated at http://oeis.org/A002109 where it says

$\log a(n) = 0.5 n^2 \log n - n^2/4 + O(n \log n)$ [Charles R Greathouse IV, Jan 12 2012] - corrected by Vaclav Kotesovec, Feb 20 2015.

and also

$a(n) \sim A n^{n(n+1)/2 + 1/12} / \exp(n^2/4)$, where $A = 1.2824271291006226368753425\dots$ is the Glaisher-Kinkelin constant (see A074962) - Vaclav Kotesovec, Feb 20 2015.