$ \log(n!) = \log(1) + \ldots + \log(n) $. This sum is $ \int_{1}^{2} \log(2) + \ldots + \int_{n-1}^{n} \log(n) $ $ \geq \int_{1}^{2} \log(t)dt + \ldots + \int_{n-1}^{n} \log(t)dt $ $ = \int_{1}^{n} \log(t) dt = n \log(n) - n +1 $. Also $ \log(2) + \ldots + \log(n-1) $ $ = \int_{2}^{3} \log(2) + \ldots + \int_{n-1}^{n} \log(n-1) $ $ \leq \int_... Read more 07 Aug 2021 - less than 1 minute read
[This is a weaker version, but the proof comes naturally] Consider $ \mathbb{R}^n $. We know $\lVert x\rVert_2 := \sqrt{\sum x^2_i} $ is a norm (ie “behaves like length”, especially $ \lVert x+y\rVert_2 \leq \lVert x\rVert_2 + \lVert y\rVert_2 $). So we can ask ourselves : $ \color{goldenrod}{\text{Q}} $) For integer $ p\geq 1$, taking $ \lVer... Read more 07 Aug 2021 - 3 minute read
$ \color{goldenrod}{\text{Th}} $: Consider sequence $ (a_j) $ in $ \mathbb{R}_ {\geq 0}$. Now $ \displaystyle \sum_{j=1}^{\infty} a_j $ converges if and only if $ \displaystyle \sum_{j=1}^{\infty} \dfrac{a_j}{1+a_j} $ converges. $ \color{goldenrod}{\text{Pf}}$: $ \underline{\implies} $ Clear as $ \frac{a_j}{1+a_j} \leq a_j $. $ \underline{\im... Read more 06 Aug 2021 - 1 minute read
$ \color{goldenrod}{\text{Th}} $: Consider series $ \sum_{n=1}^{\infty} v_n $, where $ (v_n) $ is a seq in a complete normed space. Let $ \ell $ $ := \limsup_{n\to\infty} |v_n|^{\frac{1}{n}} \in [0, \infty] $. Now : If $ 1 < \ell \leq \infty $, the series diverges. If $ \ell < 1 $, the series converges absolutely. $ \color{goldenrod}{\... Read more 06 Aug 2021 - 5 minute read
$ \color{goldenrod}{\text{Th}} $: For any $ \sigma \in S_n $, we have $ \text{ord}(\sigma) \leq (e^{\frac{1}{e}})^n \text{ } ( \leq 1.445^n ). $ $ \color{goldenrod}{\text{Pf}} $: Let $ \sigma = \sigma_1 \ldots \sigma_k $ be decomposition into disjoint cycles, and $ | \sigma_i | =: \ell_i $. Now $ \text{ord}(\sigma) $ $ = \text{LCM}(\ell_1, \ldo... Read more 06 Aug 2021 - 1 minute read