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Homeproof of alternating series test

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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. From above, we have the following:

$\displaystyle S_{{2n+1}}$ | $\displaystyle=S_{{2n}}+a_{{2n+1}};$ | ||

$\displaystyle S_{{2n+2}}$ | $\displaystyle=S_{{2n}}+(a_{{2n+1}}-a_{{2n+2}});$ | ||

$\displaystyle S_{{2n+3}}$ | $\displaystyle=S_{{2n+1}}-(a_{{2n+2}}-a_{{2n+3}})$ | ||

$\displaystyle=S_{{2n+2}}+a_{{2n+3}}$ |

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.

## Mathematics Subject Classification

40A05*no label found*

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## Corrections

grammar nazi by djao ✓

remaining minor points by djao ✓

a_i instead of a_1 by Wkbj79 ✓

consistency and grammar by Wkbj79 ✓