Post History
#2: Post edited
by
Snoopy
·
2024-08-11T16:31:13Z (3 months ago)
- If unfamiliar with Stirling's formula, one can still get a rough approximation for $n!$ using elementary calculations.
- The basic idea is to consider $\log(n!)$, which converts products into a sum that can then be analyzed using integrals. Observe that:
- $$
- \log(n!) = \sum_{m=1}^n\log m
- $$
- This sum is a Riemann sum approximation to the integral $\displaystyle \int_1^n\log(x)\ dx$:
- $$
- \int_1^n\log(x)\ dx \le \sum_{m=1}^n\log m \le \log(n) + \int_1^n\log(x)\ dx
- $$
- By finding an antiderivative for $\log(x)$ and applying the fundamental theorem of calculus, and then rearranging, one gets:
- $$
en^ne^{-n} \le n! \le en \cdot n^n e^{-n}- $$
- This leads to:
- $$
- \frac{n^n}{n!}e^{-n} \ge \frac{1}{en}
- $$
Given the divergence of the harmonic series and comparison test, we can conclude that the series diverges.
- If unfamiliar with Stirling's formula, one can still get a rough approximation for $n!$ using elementary calculations.
- The basic idea is to consider $\log(n!)$, which converts products into a sum that can then be analyzed using integrals. Observe that:
- $$
- \log(n!) = \sum_{m=1}^n\log m
- $$
- This sum is a Riemann sum approximation to the integral $\displaystyle \int_1^n\log(x)\ dx$:
- $$
- \int_1^n\log(x)\ dx \le \sum_{m=1}^n\log m \le \log(n) + \int_1^n\log(x)\ dx
- $$
- By finding an antiderivative for $\log(x)$ and applying the fundamental theorem of calculus, and then rearranging, one gets:
- $$
- en^ne^{-n} \le n! \le en \cdot n^n e^{-n}\tag{*}
- $$
- This leads to:
- $$
- \frac{n^n}{n!}e^{-n} \ge \frac{1}{en}
- $$
- Given the divergence of the harmonic series and comparison test, we can conclude that the series diverges.
- **Note**: the lower bound for $n!$ is unnecessary for this problem. Stirling's formula shows that $n!$ actually lies roughly at the [geometric mean](https://en.wikipedia.org/wiki/Geometric_mean) of the two bounds in (*).
#1: Initial revision
by
Snoopy
·
2024-08-11T16:24:56Z (3 months ago)
If unfamiliar with Stirling's formula, one can still get a rough approximation for $n!$ using elementary calculations.
The basic idea is to consider $\log(n!)$, which converts products into a sum that can then be analyzed using integrals. Observe that:
$$
\log(n!) = \sum_{m=1}^n\log m
$$
This sum is a Riemann sum approximation to the integral $\displaystyle \int_1^n\log(x)\ dx$:
$$
\int_1^n\log(x)\ dx \le \sum_{m=1}^n\log m \le \log(n) + \int_1^n\log(x)\ dx
$$
By finding an antiderivative for $\log(x)$ and applying the fundamental theorem of calculus, and then rearranging, one gets:
$$
en^ne^{-n} \le n! \le en \cdot n^n e^{-n}
$$
This leads to:
$$
\frac{n^n}{n!}e^{-n} \ge \frac{1}{en}
$$
Given the divergence of the harmonic series and comparison test, we can conclude that the series diverges.