Remainder of a geometric series ($R_N=a{r}^N/(1-r)$)

For $|r|<1$, let $S_N$ contain the first $N$ terms, with $S_0=0$. The remainder after those terms isThis is the exact absolute error. A convenient bound replaces the denominator by $1-|r|$, giving $|R_N|\le |a||r{|}^N/(1-|r|)$ through the triangle inequality. The exact denominator is preferable when the sign or complex phase of $r$ is known.

For $a>0$ and $0<r<1$, the omitted fraction of the entire sum is $R_N/(a/(1-r))=r^N$. Capturing at least $99$ percent therefore means $r^N\le 0.01$. Taking logarithms and applying the ceiling function gives the smallest integer $N$:This gives $7,21,90,$ and $459$ terms for $r=0.5,0.8,0.95,$ and $0.99$, respectively. The difference is a concrete reason to care about the common ratio, even when all four series converge.

An absolute target has a different scale. For $a=1$, $r=0.8$, and required absolute error at most $0.01$, the total sum is $5$ andThe first acceptable term count is $N=28$: the remainder is approximately $0.0096714$, while at $N=27$ it is approximately $0.0120893$. Counting terms and checking the tail avoids an ambiguous instruction to stop when a term merely looks small.