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proof of absolute convergence theorem
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(Proof)
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Suppose that $\sum a_n$ is absolutely convergent, i.e., that $\sum \vert a_n \vert$ is convergent. First of all, notice that $$ 0\le a_n + \vert a_n \vert \le 2 \vert a_n\vert, $$ and since the series $\sum (a_n+\vert a_n\vert)$ has non-negative terms it can be compared with $\sum 2\vert a_n\vert=2\sum \vert a_n\vert$ and hence converges.
On the other hand $$ \sum_{n=1}^N a_n = \sum_{n=1}^N (a_n+\vert a_n\vert) - \sum_{n=1}^N \vert a_n\vert. $$ Since both the partial sums on the right hand side are convergent, the partial sum on the left hand side is also convergent. So, the series $\sum a_n$ is convergent.
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"proof of absolute convergence theorem" is owned by paolini.
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Cross-references: left hand side, right hand side, partial sums, converges, terms, series, convergent, absolutely convergent
This is version 3 of proof of absolute convergence theorem, born on 2003-06-16, modified 2003-06-20.
Object id is 4373, canonical name is ProofOfAbsoluteConvergenceTheorem.
Accessed 2941 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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