Nannanchan math notes

Problem 1. Prove that $$\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}}{\sqrt{i^2+j^2+k^2}}=12\pi \sum_{\substack{m,n=1\\ m,n\;\mathrm{odd}}}^{\infty} \mathrm{sech}^2\bigg(\frac{\pi}{2}\sqrt{m^2+n^2}\bigg)$$ where the prime indicates that the pole $(i,j,k)=(0,0,0)$ is left out. Solution 1. It follows by symmetry that $$\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}}{\sqrt{i^2+j^2+k^2}} = \sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}(i^2+j^2+k^2)}{\sqrt{(i^2+j^2+k^2)^3}}\\ = 3\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^{i+j+k}\cdot i^2}{\sqrt{(i^2+j^2+k^2)^3}}\\ = 3\sideset{}{'}\sum_{i,j,k=-\infty}^{\infty} \frac{(-1)^i i^2(-1)^{j+k}}{\sqrt{(i^2+j^2+k^2)^3}}$$ Now using the Mellin transform $$\mathcal{M}\{f(x)\}(s)=\int_0^{\infty} x^{s-1}f(x)\,dx$$ Notice that for $p>0$, we have $$\mathcal{M}\{e^{-px}\}(s)=\int_0^{\infty} x^{s-1}e^{-px}\,dx=\int_0^{\infty} \bigg(\frac{u}{p}\bigg)^{s-1}e^{-u}\,\frac{du}{p}...

Theorem 1 (Dirichlet's Approximation Theorem). If $\alpha\in\mathbb{R}$, $t\in\mathbb{R}$, and $t\geq1$, then there exist $p\in\mathbb{Z}$ and $q\in\mathbb{N}$ such that \begin{align} \bigg\lvert \alpha-\frac{p}{q} \bigg\rvert<\frac{1}{qt}, \quad 0<q\leq t \end{align} Proof of Theorem 1. Let $\{a\}$ and $[a]$ denote the fractional part and the integer part of $a$ respectively. We set $T=[t]+1$ and consider the $T+1$ numbers: \begin{align} 0=\{\alpha\cdot0\},\{\alpha\cdot1\},\dots,\{\alpha(T-1)\},1 \end{align} where $$\{\alpha x\}=\alpha x-[\alpha x], \quad 0\leq \{\alpha x\}<1, \quad x=0,1,\dots,T-1$$ All of the points belong to the interval $0\leq y\leq 1$. We divide this interval into $T$ equal parts of the length $\frac{1}{T}$: \begin{align} \frac{k}{T}\leq y< \frac{k+1}{T}, \quad k=0,1,\dots,T-2, \quad \frac{T-1}{T}\leq y\leq 1 \end{align} Each of the points lies in one of the intervals. Since the number of points, namely $T+1$ points, is greater than the numbe...