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I played with the natural logarithm (injectivity) and ended with a Riemann sum. When n is converging against infinity, the sum converges against -1:

$$\int_{0}^1 \log(x) \, \mathrm{d} x = \lim_{x\to 0} 1 \cdot \log(1)-1 -\left( x \log(x)-x\right)$$

I do not know if that is correct. Please help!

Arturo Magidin
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2 Answers2

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Using ratio test and root test for $\sum_{n=1}^\infty a_n$ where $a_n:=\frac{n^n}{n!} > 0$, we have $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = q => \lim_{n\to\infty} \sqrt[n]{a_n} = q$$ Therefore $$\frac{a_{n+1}}{a_n} = \frac{(n+1)\cdot(n+1)^n}{(n+1)\cdot n^n} = (1+\frac{1}{n})^n \to e$$ and $$\lim_{n\to\infty} \sqrt[n]{a_n} = \lim_{n\to\infty} \sqrt[n]{\frac{n^n}{n!}} = e$$ hence result.

ThomasL
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Taking the logarithm leads to $$ \frac{1}{n} \ln (n!) - \ln n = \frac{1}{n} \sum_{k=1}^n \ln k - \ln n = \frac{1}{n}\sum_{k=1}^n \ln \frac{k}{n} \longrightarrow \int_0^1 \ln x \, dx. $$ It's easy to see (using integration by parts) that $$ \int_0^1 \ln x \, dx = x\ln x \big|_0^1 - \int_0^1 \, dx = -1. $$

This means that $$ \sqrt[n]{\frac{n!}{n^n}} = \exp \left( \frac{1}{n} \ln (n!) - \ln n\right) \to e^{-1}. $$

Edit:

I'll also add some comments on why we have the convergence $$ \frac{1}{n}\sum_{k=1}^n \ln \frac{k}{n} \longrightarrow \int_0^1 \ln x \, dx. $$ As @DanielSchepler pointed out in comments this fact is non-trivial since the right-hand-side integral is an improper one. Summing simple inequalities $$ \frac{1}{n} \ln \frac{k}{n} \le \int_{\frac{k}{n}}^{\frac{k+1}{n}} \ln x \, dx \le \frac{1}{n} \ln \frac{k+1}{n} $$ one can get $$ \frac{1}{n} \sum_{k=1}^{n-1} \ln \frac{k}{n} \le \int_{\frac{1}{n}}^{1} \ln x \, dx \le \frac{1}{n} \sum_{k=2}^{n} \ln \frac{k}{n}. $$ Therefore, $$ \int_{\frac{1}{n}}^1 \ln x \, dx + \frac{1}{n} \ln \frac{1}{n} \le \frac{1}{n} \sum_{k=1}^n \ln \frac{k}{n} \le \int_{\frac{1}{n}}^1 \ln x \, dx. $$ This gives a necessary convergence. Of course, the same procedure can be applied to any monotone function with convergent improper integral.

Virtuoz
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  • There are some subtleties here given that $\ln x$ has a singularity at 0, so the integral is an improper integral and differently chosen Riemann sums don't necessarily converge to the value of the improper integral. You would need to argue why in this case the right-hand Riemann sums do converge to the improper integral. – Daniel Schepler Apr 22 '21 at 18:30
  • @DanielSchepler indeed, you are absolutely right. I've added the clarification on convergence to my answer. – Virtuoz Apr 22 '21 at 19:04
  • @DanielSchepler For any $0<\varepsilon>0$, write $$\lim_{n\to \infty}\frac{1-\varepsilon}{n}\sum_{i=1}^n \log\left(\varepsilon+\frac in (1-\varepsilon)\right)=\int_\varepsilon^1 \log(x),dx$$. Then, let $\varepsilon\to0^+$. – Mark Viola Apr 22 '21 at 20:00
  • @MarkViola one needs to justify why it's possible to rearrange the limits then $$ \lim_{\varepsilon \to +0}\lim_{n\to \infty}\frac{1-\varepsilon}{n}\sum_{i=1}^n \log\left(\varepsilon+\frac in (1-\varepsilon)\right)=\lim_{n\to \infty} \lim_{\varepsilon \to +0}\frac{1-\varepsilon}{n}\sum_{i=1}^n \log\left(\varepsilon+\frac in (1-\varepsilon)\right) $$ – Virtuoz Apr 22 '21 at 20:12
  • @Virtuoz Just show that for all $\varepsilon\in (0,1)$ $$\lim_{n\to\infty}\frac1n \sum_{i=1}^n \log\left(1+ \frac{\varepsilon n}{i(1-\varepsilon)} \right)=0$$ – Mark Viola Apr 22 '21 at 20:19