Determine the limit of the sequence whose general term

Determine the limit of the sequence whose general term is: $$ a_n = \frac{1}{n^n} \lim_{A \to \infty} \int_0^A x^n e^{-x} \, dx $$

We need to evaluate the improper integral and then find the limit of the resulting sequence.

Step 1: Evaluate the improper integral

First, let's evaluate:

$$\int_0^\infty x^n e^{-x} \, dx$$

This is a standard integral related to the Gamma function.

Recall the Gamma function:

$$\Gamma(n) = \int_0^\infty t^{n-1} e^{-t} \, dt \quad \text{for } n > 0$$

For our integral, substituting $n$ with $n+1$:

$$\int_0^\infty x^n e^{-x} \, dx = \Gamma(n+1)$$

Property of the Gamma function:

For positive integers $n$:

$$\Gamma(n+1) = n!$$

Therefore:

$$\int_0^\infty x^n e^{-x} \, dx = n!$$

Step 2: Express the general term

Now we have:

$$a_n = \frac{1}{n^n} \cdot n! = \frac{n!}{n^n}$$

Step 3: Find $\lim_{n \to \infty} a_n$

We need to find:

$$\lim_{n \to \infty} \frac{n!}{n^n}$$

Using Stirling's approximation:

Stirling's formula states that for large $n$:

$$n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$$

More precisely:

$$\lim_{n \to \infty} \frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n} = 1$$

Therefore:

$$a_n = \frac{n!}{n^n} = \frac{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{n^n} \cdot \frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}$$

$$= \frac{\sqrt{2\pi n} \cdot n^n}{e^n \cdot n^n} \cdot \frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}$$

$$= \frac{\sqrt{2\pi n}}{e^n} \cdot \frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}$$

As $n \to \infty$, the second factor approaches 1, so:

$$a_n \sim \frac{\sqrt{2\pi n}}{e^n}$$

Therefore:

$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt{2\pi n}}{e^n} = 0$$

Since $e^n$ grows exponentially while $\sqrt{n}$ grows polynomially, the limit is 0.

Alternative approach: Direct analysis

We can also see this directly. We have:

$$a_n = \frac{n!}{n^n} = \frac{n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1}{n \cdot n \cdot n \cdots n \cdot n}$$

$$= \frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{2}{n} \cdot \frac{1}{n}$$

$$= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{2}{n} \cdot \frac{1}{n}$$

Each factor $\frac{k}{n}$ for $k < n$ is less than 1. We can write:

$$a_n = \prod_{k=1}^n \frac{k}{n} = \prod_{k=1}^n \left(1 - \frac{n-k}{n}\right)$$

Taking logarithms:

$$\ln a_n = \sum_{k=1}^n \ln\frac{k}{n} = \sum_{k=1}^n \ln k - n \ln n$$

$$= \ln(n!) - n\ln n$$

Using Stirling's approximation $\ln(n!) \approx n\ln n - n + \frac{1}{2}\ln(2\pi n)$:

$$\ln a_n \approx (n\ln n - n + \frac{1}{2}\ln(2\pi n)) - n\ln n$$

$$= -n + \frac{1}{2}\ln(2\pi n)$$

$$= -n + \frac{1}{2}\ln(2\pi) + \frac{1}{2}\ln n$$

As $n \to \infty$, $-n$ dominates, so $\ln a_n \to -\infty$.

Therefore:

$$a_n = e^{\ln a_n} \to e^{-\infty} = 0$$

Final Answer

$$\boxed{\lim_{n \to \infty} a_n = 0}$$

Rate of convergence

From Stirling's approximation, we have:

$$a_n \sim \frac{\sqrt{2\pi n}}{e^n}$$

This shows that $a_n$ converges to 0 exponentially fast, with the decay rate dominated by $e^{-n}$.

More precisely, for large $n$:

$$a_n \approx \sqrt{2\pi n} \cdot e^{-n}$$

Numerical verification

Let's check a few values:

For $n = 1$:

$$a_1 = \frac{1!}{1^1} = \frac{1}{1} = 1$$

For $n = 2$:

$$a_2 = \frac{2!}{2^2} = \frac{2}{4} = 0.5$$

For $n = 3$:

$$a_3 = \frac{3!}{3^3} = \frac{6}{27} \approx 0.222$$

For $n = 4$:

$$a_4 = \frac{4!}{4^4} = \frac{24}{256} = 0.09375$$

For $n = 5$:

$$a_5 = \frac{5!}{5^5} = \frac{120}{3125} = 0.0384$$

For $n = 10$:

$$a_{10} = \frac{10!}{10^{10}} = \frac{3628800}{10000000000} \approx 0.00036288$$

We can see the sequence is rapidly decreasing toward 0.

Comparison with Stirling's approximation

For $n = 10$:

$$a_{10} \approx \frac{\sqrt{2\pi \cdot 10}}{e^{10}} \approx \frac{\sqrt{62.83}}{22026.47} \approx \frac{7.926}{22026.47} \approx 0.00036$$

This matches our exact calculation very well!

Interpretation

The sequence $a_n = \frac{n!}{n^n}$ can be interpreted as:

$$a_n = \frac{\text{number of permutations of } n \text{ objects}}{\text{number of } n\text{-tuples with repetition}}$$

Alternatively, if we select $n$ items from $n$ choices with replacement, there are $n^n$ possible outcomes. If we select without replacement and care about order, there are $n!$ outcomes. The ratio $\frac{n!}{n^n}$ represents the probability that all $n$ selections are distinct when sampling with replacement.

As $n$ increases, this probability decreases to 0 (it becomes increasingly unlikely that all selections are distinct).

Summary

1. We recognized that $\int_0^\infty x^n e^{-x} \, dx = \Gamma(n+1) = n!$
2. This gave us $a_n = \frac{n!}{n^n}$
3. Using Stirling's approximation: $$ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n $$
4. We found: $$ a_n \sim \frac{\sqrt{2\pi n}}{e^n} \to 0 $$
5. The sequence converges to 0 exponentially fast

Key lessons:

• The Gamma function $\Gamma(n+1) = n!$ connects improper integrals to factorials
• Stirling's approximation is essential for analyzing factorials in limits
• The ratio $\frac{n!}{n^n}$ decreases exponentially to 0
• This has a probabilistic interpretation related to the "birthday paradox" type problems
• Exponential decay dominates polynomial growth in limits