= Reciprocal-factorial series remainder bound
{title2=$e-\sum_{i=0}^n\frac1{i!}\leq\frac{n+2}{(n+1)(n+1)!}$}
The positive tail after term $n$ in the reciprocal-<factorial> series for <Euler's number> satisfies $1/(n+1)!<e-\sum_{i=0}^n1/i!\leq(n+2)/((n+1)(n+1)!)$. The upper bound replaces each subsequent denominator factor by $n+2$ and sums the resulting <geometric series>; the lower bound retains only the first omitted term. This supplies an explicit error certificate and, with <Stirling's formula>, a term count of order $d/\log d$ for $d$ decimal digits. The arithmetic evaluation also has its own rounding error, distinct from this truncation bound.
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