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HomeLeibniz' estimate for alternating series

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Theorem (Leibniz 1682). If $p_{1}>p_{2}>p_{3}>\cdots$ and $\displaystyle\lim_{{m\to\infty}}p_{m}=0$, then the alternating series

$\displaystyle p_{1}-p_{2}+p_{3}-p_{4}+-\ldots$ | (1) |

converges. Its remainder term has the same sign as the first omitted term $\pm p_{{m+1}}$ and the absolute value^{} less than $p_{{m+1}}$.

Proof. The convergence of (1) is proved here. Now denote the sum of the series by $S$ and the partial sums of it by $S_{1},\,S_{2},\,S_{3},\,\ldots$. Suppose that (1) is truncated after a negative term $-p_{{2n}}$. Then the remainder term

$R_{{2n}}\;=\;S\!-\!S_{{2n}}$ |

may be written in the form

$R_{{2n}}\;=\;(p_{{2n+1}}-p_{{2n+2}})+(p_{{2n+3}}-p_{{2n+4}})+\ldots$ |

or

$R_{{2n}}\;=\;p_{{2n+1}}-(p_{{2n+2}}-p_{{2n+3}})-(p_{{2n+4}}-p_{{2n+5}})-\ldots$ |

The former shows that $R_{{2n}}$ is positive as the first omitted term $p_{{2n+1}}$ and the latter that $|R_{{2n}}|<p_{{2n+1}}$. Similarly one can see the assertions true when the series (1) is truncated after a positive term $p_{{2n-1}}$.

*A pictorial proof.*

As seen in this diagram, whenever $m^{{\prime}}>m$, we have
$\abs{S_{{m^{{\prime}}}}\!-\!S_{m}}\leq p_{{m+1}}\to 0$. Thus the partial sums form a Cauchy sequence^{}, and hence converge. The limit lies in the centre of the spiral, strictly in between $S_{m}$ and $S_{{m+1}}$ for any $m$. So the remainder after the $m$th term must have the same direction as $\pm p_{{m+1}}=S_{{m+1}}\!-\!S_{m}$ and lesser magnitude.

Example 1. The alternating series

$\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{% 3}+1}+\frac{1}{\sqrt{4}-1}-\frac{1}{\sqrt{4}+1}+-\ldots$ |

does not fulfil the requirements of the theorem and is divergent.

Example 2. The alternating series

$\frac{1}{\ln{2}}-\frac{1}{\ln{3}}+\frac{1}{\ln{4}}-\frac{1}{\ln{5}}+-\ldots$ |

satisfies all conditions of the theorem and is convergent.

## Mathematics Subject Classification

40A05*no label found*40-00

*no label found*

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