Evaluate $\lim\limits_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}.$ Suppose $\alpha>1$, and $a_n$ be the largest natrual number $k$ such that $k!\le \alpha^n$.Evaluate $$\lim\limits_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}.$$
According to the assumption, we obtain
$$(a_n)!\le \alpha^n<(a_n+1)!=(a_n+1)(a_n)!.$$
Therefore
$$ (a_1)!(a_2)!\cdots(a_n)!\le \alpha^{1+2+\cdots+n}<(a_1+1)(a_2+1)\cdots(a_n+1)(a_1)!(a_2)!\cdots(a_n)!.$$
But how to go on?
 A: I guess answer is 0.5!
Here's some lemma I could prove:
Lemma 1: $\{a_n\}_{n=1}^{\infty}$ is increasing. (obviously)
Let $a_{t-1} \leq \alpha \leq a_t $.
Lemma 2: for $i\geq t, a_{i+1} = \ a_i$ or $a_{i+1}= a_i + 1$.
Proof: if $a_i = a_{i+1}=...=a_{i+r}\leq a_{i+r+1}$ then
$$a_{i+r}!\leq \alpha ^{i+r-1} \leq (a_{i+r}+1)!$$
$$a_{i+r+1} !\leq \alpha ^{i+r}\leq (a_{i+r+1}+1)!$$
so $\alpha\ a_{i+r}!\leq \alpha ^{i+r} \leq \alpha\ (a_{i+r}+1)!$ then $a_{i+r+1}!\leq \alpha\ (a_{i+r}+1)! \leq a_i \ (a_{i+r}+1)!$. Now if $a_{i+r+1} >a_{i+r}+1$ we come to a contradiction by last inequality!!! So $a_i\leq a_{i+1}\leq a_i +1$ for $i\geq t.$
Lemma 3: by Stirling's approximation $a_n!\approx \sqrt{2\pi a_n}(\frac{a_n}{e})^{a_n}\leq \alpha^n\Rightarrow a_n \ln a_n \leq n \ln \alpha\Rightarrow \frac{\ln a_n}{\ln \alpha} \leq \frac{n}{a_n}\Rightarrow \lim_{n\to \infty}\frac{n}{a_n}=\infty\Rightarrow a_n\leq n$

Lemma 4: let $A_n = \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n} $ then for large enough $n, \lim_{n\to\infty} A_n < \infty.$
Proof:
$$\lim_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}\ =\ \lim_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_{t-1}a_t \cdots a_n}}{a_n}=
 \lim_{n \to \infty} \frac{\sqrt[n]{a_ta_t\cdots a_t\ (a_t+1)(a_t+2) \cdots (a_t+n-t)}}{a_n}< \lim_{n \to \infty} \frac{a_t+a_t+\cdots +a_t+ (a_t+1)+(a_t+2)+ \cdots +(a_t+n-t)}{na_n}=\lim_{n \to \infty} \frac{na_t+1+2+\cdots (n-t)}{na_n}=\lim_{n \to \infty} \frac{a_t}{a_n}+\frac{(n-t+1)(n-t)}{2na_n}=0+\lim_{n \to \infty} \frac{n}{a_n}=\infty$$


Finally by running a programm in matlab:\
figure;
for j=0:1:10
alpha = 1+j;
A = [];
B = [];
for i=1:100
    k = 1;
    while (factorial(k) <= power(alpha, i))
        k = k + 1;
    end
    A = [A, k-1];
    B = [B, nthroot(prod(A),i)/A(end)];
end

%display(A);
%display(B);
%figure;
%plot(A);
%hold on;
%figure;
plot(B);
hold on;
end

limit of A_n

limit of a_n

A: Hint:  When $n = 1$, $\ln \sqrt{a_1/a_1} = 0$.
When $n = 2$,
$$  \ln \sqrt{\frac{a_1}{a_2} \cdot \frac{a_2}{a_2}} = \frac{1}{2}\left( \ln(a_1) - \ln(a_2) + 0 \right) < 0  \text{.}  $$
Now, $\log_\alpha (a_j) \ll \log_\alpha (a_j!) \leq \log_\alpha (\alpha^j) = j$ for all $j$.  Can you get a sharper version of "$\ll$"?
