= Tail of a telescoping reciprocal series
{title2=$\sum_{n=N+1}^{\infty}\frac1{n(n+1)}=\frac1{N+1}$}
For the <telescoping series> $\sum_{n=1}^{\infty}1/[n(n+1)]=1$, the remainder after the first $N$ terms is exactly
$$
1-S_N=\sum_{n=N+1}^{\infty}\frac1{n(n+1)}=\frac1{N+1}.
$$
Thus $S_9=0.9$, $S_{99}=0.99$, and $S_{999}=0.999$. This supplies an exact stopping rule: to make the <absolute error> at most $10^{-3}$, \b[999 terms suffice]. Requiring strictly less than $10^{-3}$ instead needs at least 1000 terms.
The next term, $a_{N+1}=1/[(N+1)(N+2)]$, is smaller than the full remaining <sum> by a factor of $N+2$. For example, after 99 terms the next term is $1/10100$, while the remainder is $1/100$. The <term test for divergence> concerns individual terms; a reliable numerical approximation needs a bound on the whole remaining <sum>.
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