proof of Leibniz’s theorem (using Dirichlet’s convergence test)

Proof.Let us define the sequence ${\alpha }_n={(-1)}^n$ for$n\in \mathbb{N}=\{0,1,2,\mathrm{\dots }\}.$ Then

so the sequence ${\sum }_{i=0}^n{\alpha }_i$ is bounded.By assumption ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ is a bounded decreasingsequence with limit $0$.For $n\in \mathbb{N}$ we set $b_n:=a_{n+1}$.Using Dirichlet’s convergence test, it follows that the series${\sum }_{i=0}^{\mathrm{\infty }}{\alpha }_i{b}_i$ converges. Since

$$\sum _{i=0}^{\mathrm{\infty }}{\alpha }_i{b}_i=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n+1}{a}_n,$$

the claim follows.$\mathrm{\square }$