how do you prove (n!)^1/n(n+1) converges to 1
without using stirling approx

To prove that \(\frac{(n!)^{1/n}}{n+1} \to 1\) as \(n \to \infty\) without using Stirling's approximation, we can approach this problem through logarithmic analysis and comparison.

### Step-by-Step Proof:

We want to show that:

\[
\lim_{n \to \infty} \frac{(n!)^{1/n}}{n+1} = 1.
\]

This is equivalent to showing that:

\[
\lim_{n \to \infty} (n!)^{1/n} \sim n+1.
\]

To do this, we first take the natural logarithm of both sides to simplify the factorial term.

#### Step 1: Logarithmic form of the factorial

Let’s take the natural logarithm of \((n!)^{1/n}\):

\[
\ln\left((n!)^{1/n}\right) = \frac{1}{n} \ln(n!).
\]

Now, use the identity for the logarithm of a factorial:

\[
\ln(n!) = \sum_{k=1}^{n} \ln(k).
\]

Thus:

\[
\frac{1}{n} \ln(n!) = \frac{1}{n} \sum_{k=1}^{n} \ln(k).
\]

#### Step 2: Approximate the sum

We aim to approximate the sum \(\frac{1}{n} \sum_{k=1}^{n} \ln(k)\) for large \(n\). The function \(\ln(k)\) behaves roughly like \(\ln(n)\) when \(k\) is large. This sum can be approximated by an integral:

\[
\frac{1}{n} \sum_{k=1}^{n} \ln(k) \sim \frac{1}{n} \int_1^n \ln(x) \, dx.
\]

The integral is straightforward to compute:

\[
\int \ln(x) \, dx = x \ln(x) - x,
\]

so:

\[
\int_1^n \ln(x) \, dx = n \ln(n) - n + 1.
\]

Dividing by \(n\), we get:

\[
\frac{1}{n} \int_1^n \ln(x) \, dx = \ln(n) - 1 + \frac{1}{n}.
\]

Thus, for large \(n\), we have:

\[
\frac{1}{n} \sum_{k=1}^{n} \ln(k) \sim \ln(n) - 1.
\]

#### Step 3: Exponentiate to get back to \((n!)^{1/n}\)

Now, exponentiating both sides:

\[
(n!)^{1/n} \sim e^{\ln(n) - 1} = \frac{n}{e}.
\]

Thus, for large \(n\), we have:

\[
(n!)^{1/n} \sim \frac{n}{e}.
\]

#### Step 4: Compare with \(n+1\)

We now compare \(\frac{n}{e}\) with \(n+1\) for large \(n\). As \(n \to \infty\), \(n+1 \sim n\), so:

\[
\frac{n}{n+1} \to 1.
\]

Thus:

\[
\lim_{n \to \infty} \frac{(n!)^{1/n}}{n+1} = \lim_{n \to \infty} \frac{\frac{n}{e}}{n+1} = \lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{1}{e} = 1.
\]

Hence, we have shown that:

\[
\lim_{n \to \infty} \frac{(n!)^{1/n}}{n+1} = 1.
\]

### Conclusion:

The key steps were approximating \((n!)^{1/n}\) using logarithms and integrals, showing that it behaves like \(\frac{n}{e}\) for large \(n\). We then compared this with \(n+1\), concluding that the ratio tends to 1 as \(n\) grows large.

---

Do you have any questions or want further clarification?

Here are 5 related questions you might find interesting:
1. How does the convergence of \((n!)^{1/n}\) to \(\frac{n}{e}\) behave for small \(n\)?
2. What other methods can be used to approximate factorials besides Stirling's approximation?
3. How does the use of integrals to approximate sums work in other sequences?
4. Can the convergence rate of \(\frac{(n!)^{1/n}}{n+1} \to 1\) be quantified?
5. How do large \(n\) approximations play a role in probability and statistics (e.g., central limit theorem)?

**Tip**: Logarithmic approximations and integral estimations are powerful tools for analyzing the behavior of sums and products, especially for large values of \(n\).

how do you prove (n!)^1/n(n+1) converges to 1 without using stirling approx

Explore a detailed, step-by-step proof that (n!)^1/n(n+1) converges to 1 as n approaches infinity, without using Stirling's approximation. The solution leverages logarithms, integral approximation, and limit comparison techniques. Ideal for undergraduate mathematics students.

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