A Brilliant Limit
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it as the source of the Riemann-sum trick for this limit, both when asking follow-up questions about its steps and when explaining how sums become integrals.
Recommended by 1 people across 1 different months, 2018 to 2022.
What it is
blackpenredpen works through evaluating the limit of (n!/n^n)^(1/n) as n goes to infinity. He takes the log, turns the resulting sum into a Riemann sum, and evaluates the integral of ln(x) from 0 to 1 to get 1/e. It is a single whiteboard-style calculus walkthrough.
What people said
Recommends thewinster · 2022-05 · Mathematics Stack Exchange · on “Which sum converts into a definite integral? $\sum_{i=0}^n \frac {e ^ i}{n}$ or $\sum_{i=0}^n \frac {e ^ \frac{i}{n}}{n}$”
Which sum converts into a definite integral? $\sum_{i=0}^n \frac {e ^ i}{n}$ or $\sum_{i=0}^n \frac {e ^ \frac{i}{n}}{n}$ Or none of the two do? I have come across problems such as $ \lim_{n_\to infinity} (\frac{n!}{n^n})^\frac{1}{n}$ . When we take a log of this limit, the product of n! and n.n.n. ... n turns into a sum, while the exponent 1/n becomes a common term for the entire sum. This expression now becomes a definite integral from 0 to 1, as explained in [this video] . I suspect that one of the two above mentioned sums, because of the presence of 1/n as a term, can turn into a definite integral. Is this true? If so, is it the sum with $e^i$ or $e^\frac{i}{n}$ ? I don't think $e^i$ sum can
Cites Aaryan Dewan · 2018-12 · Mathematics Stack Exchange · on “How do I calculate $\lim_{x\to \infty} \sqrt{(2\pi x)}^\frac{1}{x}$”
I just started studying limits a week ago and today I got this question on my YouTube feed. I am having a hard time with it. The question is: Find the value of: $\lim_{x\to \infty}(\frac{x!}{x^x})^\frac{1}{x}$ The answer to this question is $\frac{1}{e}$ There in the comment section, someone suggested to use the Stirling's approximation. It states $n! \approx e^{-n}*n^n*\sqrt{2\pi n}$ After putting this in the question it reduces down to: proving that $\lim_{x\to \infty} \sqrt{(2\pi x)}^\frac{1}{x}=1$ I just don’t know how to eliminate that $\pi$ in the expression. Wolfram Alpha suggested to use Puiseux series. Apparently, the wiki page is just too advanced for me to comprehend. I would be r
Cites selector · 2021-10 · Mathematics Stack Exchange · on “Find $\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}$”
I am referring to this video on youtube [this video] which shows the determination of $$ \lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}. $$ In the video, the trick is to set $$ L=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n} $$ and then to consider $$ \ln(L)=\ln\left(\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}\right). $$ I have some problems with the following step: It is said that, since $\ln$ is a continuous function, one has $$ \ln\left(\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}\right)=\lim_{n\to\infty}\ln\left(\left(\frac{n!}{n^n}\right)^{1/n}\right) $$ Is this really true? Doesn't this step only hold if the limit exists (what is exactly the task to show)?