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proof of Leibniz's theorem (using Dirichlet's convergence test)
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(Proof)
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Proof. Let us define the sequence $\alpha_n =(-1)^n$ for $n\in \sN=\{0,1,2,\ldots\}.$ Then
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 $n\in \sN$ 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. 
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"proof of Leibniz's theorem (using Dirichlet's convergence test)" is owned by mathcam. [ full author list (2) | owner history (1) ]
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Cross-references: converges, series, Dirichlet's convergence test, limit, decreasing, bounded, sequence, proof
This is version 4 of proof of Leibniz's theorem (using Dirichlet's convergence test), born on 2003-01-18, modified 2003-04-24.
Object id is 3900, canonical name is ProofOfLeibnizsTheorem.
Accessed 9577 times total.
Classification:
| AMS MSC: | 40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences) |
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Pending Errata and Addenda
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