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proof of alternating series test
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(Proof)
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If the first term $a_1$ is positive, then the series has partial sum $$ S_{2n+2}=a_1-a_2+a_3+...-a_{2n}+a_{2n+1}-a_{2n+2}, $$ where the $a_j$ 's are all nonnegative and nonincreasing. (If the first term is negative, consider the series in the absence of the first term.) From above, we have the following:
Since $a_{2n+1} \geq a_{2n+2}\geq a_{2n+3}$ , we have $S_{2n+1}\geq S_{2n+3} \geq S_{2n+2} \geq S_{2n}$ . Moreover, $$ S_{2n+2}=a_1-(a_2-a_3)-(a_4-a_5)-\cdots-(a_{2n}-a_{2n+1})-a_{2n+2}. $$ Because the $a_j$ 's are nonincreasing, we have $S_n \geq 0$ for any $n$ . Also, $S_{2n+2} \leq S_{2n+1} \leq a_1$ . Thus, $a_1 \geq S_{2n+1} \geq S_{2n+3} \geq S_{2n+2} \geq S_{2n} \geq 0$ . Hence, the even partial sums $S_{2n}$ and the odd partial sums $S_{2n+1}$ are bounded. Also, the even partial sums $S_{2n}$ 's are monotonically nondecreasing, while the odd partial sums $S_{2n+1}$ 's are monotonically nonincreasing. Thus, the even and odd series both converge.
We note that $S_{2n+1}-S_{2n}=a_{2n+1}$ . Therefore, the sums converge to the same limit if and only if $a_n\to 0$ as $n\to\infty$ . The theorem is then established.
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"proof of alternating series test" is owned by Wkbj79. [ full author list (2) | owner history (2) ]
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Cross-references: theorem, limit, sums, converge, monotonically nondecreasing, bounded, odd, even, negative, nonincreasing, partial sum, series, positive, term
There is 1 reference to this entry.
This is version 9 of proof of alternating series test, born on 2002-05-24, modified 2007-10-05.
Object id is 2936, canonical name is ProofOfAlternatingSeriesTest2.
Accessed 6527 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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