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

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

$\sum_{i=0}^{n}\alpha_{i}=\left\{\begin{array}[]{ll}1&\mbox{for even}\,n,\\ 0&\mbox{for odd}\,n,\end{array}\right.$

so the sequence $\sum_{i=0}^{n}\alpha_{i}$ is bounded. By assumption $\{a_{n}\}_{n=1}^{\infty}$ 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 $\sum_{i=0}^{\infty}\alpha_{i}b_{i}$ converges. Since

$\sum_{i=0}^{\infty}\alpha_{i}b_{i}=\sum_{n=1}^{\infty}(-1)^{n+1}a_{n},$

the claim follows. $\Box$

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