Leibniz’ estimate for alternating series

Theorem (Leibniz 1682).  If  $p_1>p_2>p_3>\mathrm{\cdots }$  and  $\underset{m\to \mathrm{\infty }}{lim}{p}_m=0$,  then the alternating series

$p_1-p_2+p_3-p_4+-\mathrm{\dots }$(1)

converges.  Its remainder term has the same sign (http://planetmath.org/SignumFunction) as the first omitted $\pm p_{m+1}$ and the absolute value less than $p_{m+1}$.

Proof.  The convergence of (1) is provedhere (http://planetmath.org/ProofOfAlternatingSeriesTest).  Now denote the sum of the series by $S$ and the partial sums of it by $S_1,S_2,S_3,\mathrm{\dots }$.  Suppose that (1) is truncated after a negative $-p_{2n}$.  Then the remainder term

$$R_{2n}=S-S_{2n}$$

may be written in the form

$$R_{2n}=(p_{2n+1}-p_{2n+2})+(p_{2n+3}-p_{2n+4})+\mathrm{\dots }$$

or

$$R_{2n}=p_{2n+1}-(p_{2n+2}-p_{2n+3})-(p_{2n+4}-p_{2n+5})-\mathrm{\dots }$$

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

A pictorial proof.

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

Example 1.  The alternating series

$$\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{4}-1}-\frac{1}{\sqrt{4}+1}+-\mathrm{\dots }$$

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

Example 2.  The alternating series

$$\frac{1}{\ln 2}-\frac{1}{\ln 3}+\frac{1}{\ln 4}-\frac{1}{\ln 5}+-\mathrm{\dots }$$

satisfies all conditions of the theorem and is convergent.