Post History
#4: Post edited
by
Snoopy
·
2024-08-15T12:30:12Z (4 months ago)
- The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the [limit comparison test](https://en.wikipedia.org/wiki/Limit_comparison_test). The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.
Stirling's formula,- $$
- \lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1,
- $$
- provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test:
- Consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since the (multiple of) the p-series $\sum b_n$ does.
- -----
- **Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
- $$
- n! \sim C\sqrt{n} n^n e^{-n}
- $$
- where $C$ is *some* constant.
- This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.
- The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the [limit comparison test](https://en.wikipedia.org/wiki/Limit_comparison_test). The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.
- [Stirling's formula](https://en.wikipedia.org/wiki/Stirling%27s_approximation),
- $$
- \lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1,
- $$
- provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test:
- Consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since the (multiple of) the p-series $\sum b_n$ does.
- -----
- **Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
- $$
- n! \sim C\sqrt{n} n^n e^{-n}
- $$
- where $C$ is *some* constant.
- This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.
#3: Post edited
by
Snoopy
·
2024-08-11T13:28:44Z (4 months ago)
- The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the [limit comparison test](https://en.wikipedia.org/wiki/Limit_comparison_test). The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.
- Stirling's formula,
- $$
- \lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1,
- $$
provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test. For instance, consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since the (multiple of) the p-series $\sum b_n$ does.- **Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
- $$
- n! \sim C\sqrt{n} n^n e^{-n}
- $$
- where $C$ is *some* constant.
- This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.
- The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the [limit comparison test](https://en.wikipedia.org/wiki/Limit_comparison_test). The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.
- Stirling's formula,
- $$
- \lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1,
- $$
- provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test:
- Consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since the (multiple of) the p-series $\sum b_n$ does.
- -----
- **Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
- $$
- n! \sim C\sqrt{n} n^n e^{-n}
- $$
- where $C$ is *some* constant.
- This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.
#2: Post edited
by
Snoopy
·
2024-08-11T13:27:00Z (4 months ago)
- The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the [limit comparison test](https://en.wikipedia.org/wiki/Limit_comparison_test). The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.
- Stirling's formula,
- $$
- \lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1,
- $$
provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test. For instance, consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since $\sum b_n$ does.- **Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
- $$
- n! \sim C\sqrt{n} n^n e^{-n}
- $$
- where $C$ is *some* constant.
- This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.
- The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the [limit comparison test](https://en.wikipedia.org/wiki/Limit_comparison_test). The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.
- Stirling's formula,
- $$
- \lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1,
- $$
- provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test. For instance, consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since the (multiple of) the p-series $\sum b_n$ does.
- **Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
- $$
- n! \sim C\sqrt{n} n^n e^{-n}
- $$
- where $C$ is *some* constant.
- This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.
#1: Initial revision
by
Snoopy
·
2024-08-11T13:24:25Z (4 months ago)
The basic approach to determining the convergence of a series involves comparing it to a simpler series using either the direct comparison test or the [limit comparison test](https://en.wikipedia.org/wiki/Limit_comparison_test). The latter indicates when a complicated summand can be substituted with a simpler one for easier analysis.
Stirling's formula,
$$
\lim _{n \rightarrow \infty} \frac{n!}{\sqrt{2 \pi n}(n / e)^n}=1,
$$
provides crucial insight into the rapid growth of $n!$. It enables the immediate determination of convergence using the limit comparison test. For instance, consider $a_n = \frac{n^n e^{-n}}{n!}$ and $b_n = \sqrt{2\pi n}$. Applying Stirling’s formula and the limit comparison test shows that $\sum a_n$ diverges since $\sum b_n$ does.
**Note:** Stirling's formula might be overkill for this particular problem. All that's really needed is the approximation discovered by De Moivre:
$$
n! \sim C\sqrt{n} n^n e^{-n}
$$
where $C$ is *some* constant.
This answer exemplifies a technique that is impossible or very difficult for beginner calculus students to think of, let alone apply, if they do not already have it in their toolkit. Stirling's formula, which often does not appear in standard calculus textbooks, greatly simplifies the analysis of many series involving the factorial term.