= Remainder of a geometric series
{title2=$R_N=ar^N/(1-r)$}
For $|r|<1$, let $S_N$ contain the first $N$ terms, with $S_0=0$. The remainder after those terms is
$$
R_N=\sum_{j=N}^{\infty}ar^j=\frac{ar^N}{1-r},
\qquad
\left|\frac{a}{1-r}-S_N\right|
=\frac{|a|\,|r|^N}{|1-r|}.
$$
This 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.
\Image[/geometric-series-remainder-and-term-count.png]
{title=The omitted fraction and the number of terms needed to capture 99 percent of a geometric sum}
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 $r^N\le0.01$. Taking <logarithms> and applying the <ceiling function> gives the smallest <integer> $N$:
$$
N=\left\lceil\frac{\log(0.01)}{\log r}\right\rceil.
$$
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 $5$ and
$$
|R_N|=5(0.8)^N\le0.01.
$$
The 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.
Back to article page
