Evaluating the Limit of a Product Involving Factorials and Exponents

The problem at hand is to evaluate the following limit:

(L lim_{n to infty} frac{n!^{frac{1}{n}} 2!^{frac{1}{n}} cdots n^{frac{1}{n}}}{n})

The Solution Process

First, let's simplify the expression:

[ L lim_{n to infty} frac{prod_{i1}^n i^{frac{1}{n}}}{n} lim_{n to infty} left(prod_{i1}^n frac{i}{n}right)^{frac{1}{n}}. ]

We use the fact that (ln{L} lim_{n to infty} left[-ln{n} frac{1}{n} sum_{k1}^n ln{k}right]).

Notice that (sum_{k1}^n ln{k}) can be rewritten as:

Riemann Sum Interpretation

Consider the integral (int_1^2 ln{x} , dx). The expression (frac{1}{n} sum_{k1}^n ln{left(1 frac{k}{n}right)}) is a Riemann sum approximation for this integral.

Calculation of the Integral

Evaluating the integral, we get:

[ int_1^2 ln{x} , dx x ln{x} - x Bigg|_1^2 2 ln{2} - 1 lnleft(frac{4}{e}right). ]

Hence, (ln{L} lnleft(frac{4}{e}right)).

Exponentiation

Finally, (L e^{lnleft(frac{4}{e}right)} frac{4}{e}).

Formal Proof Using Limits and Sequences

Using the formal definition, we can write the sequence as:

(a_n frac{n! 2! cdots n!}{n^n}), and (x_n sqrt[n]{a_n}).

The theorem about the limit of a sequence tells us that: [ lim_{n to infty} x_n lim_{n to infty} frac{a_{n 1}}{a_n}. ]

Now, compute (frac{a_{n 1}}{a_n}), which is much simpler:

[ frac{a_{n 1}}{a_n} frac{prod_{i1}^{n 1} i^{frac{1}{n 1}}}{prod_{i1}^n i^{frac{1}{n}}} 2 prod_{i1}^n left(frac{n 1}{n} cdot frac{ni}{(n 1)i}right) 2 left(frac{n 1}{n}right)^n. ]

As (n to infty), (left(frac{n 1}{n}right)^n to e), so (frac{a_{n 1}}{a_n} to 2e).

Hence, the limit of the original sequence is:

[ L lim_{n to infty} frac{prod_{i1}^n i^{frac{1}{n}}}{n} frac{4}{e}. ]

Discussion and Conclusion

This problem involves advanced concepts in limits, Riemann integrals, and sequences. Applying these concepts, we find that the limit of the given expression is (frac{4}{e}).