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

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

may be written in the form

or

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  $\lvert S_{{m^{{\prime}}}}\!-\!S_{m}\rvert\leq p_{{m+1}}\to 0$.  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

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

Example 2.  The alternating series

satisfies all conditions of the theorem and is convergent.