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 decreasing sequence 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. $\mathrm{\square }$

Title	proof of Leibniz’s theorem (using Dirichlet’s convergence test)
Canonical name	ProofOfLeibnizsTheoremusingDirichletsConvergenceTest
Date of creation	2013-03-22 13:22:17
Last modified on	2013-03-22 13:22:17
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	7
Author	mathcam (2727)
Entry type	Proof
Classification	msc 40A05
Related topic	AlternatingSeries