# proof of alternating series test

The series has partial sum

$${S}_{2n+2}={a}_{1}-{a}_{2}+{a}_{3}-+\mathrm{\dots}-{a}_{2n}+{a}_{2n+1}-{a}_{2n+2},$$ |

where the ${a}_{j}$’s are all nonnegative and nonincreasing. From above, we have the following:

${S}_{2n+1}$ | $={S}_{2n}+{a}_{2n+1};$ | ||

${S}_{2n+2}$ | $={S}_{2n}+({a}_{2n+1}-{a}_{2n+2});$ | ||

${S}_{2n+3}$ | $={S}_{2n+1}-({a}_{2n+2}-{a}_{2n+3})$ | ||

$={S}_{2n+2}+{a}_{2n+3}$ |

Since ${a}_{2n+1}\ge {a}_{2n+2}\ge {a}_{2n+3}$, we have ${S}_{2n+1}\ge {S}_{2n+3}\ge {S}_{2n+2}\ge {S}_{2n}$. Moreover,

$${S}_{2n+2}={a}_{1}-({a}_{2}-{a}_{3})-({a}_{4}-{a}_{5})-\mathrm{\cdots}-({a}_{2n}-{a}_{2n+1})-{a}_{2n+2}.$$ |

Because the ${a}_{j}$’s are nonincreasing, we have ${S}_{n}\ge 0$ for any $n$. Also, ${S}_{2n+2}\le {S}_{2n+1}\le {a}_{1}$. Thus, ${a}_{1}\ge {S}_{2n+1}\ge {S}_{2n+3}\ge {S}_{2n+2}\ge {S}_{2n}\ge 0$. 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 $n\to \mathrm{\infty}$. The theorem is then established.

Title | proof of alternating series test |
---|---|

Canonical name | ProofOfAlternatingSeriesTest |

Date of creation | 2014-07-22 16:20:39 |

Last modified on | 2014-07-22 16:20:39 |

Owner | Wkbj79 (1863) |

Last modified by | pahio (2872) |

Numerical id | 13 |

Author | Wkbj79 (2872) |

Entry type | Proof |

Classification | msc 40A05 |