Reciprocal-factorial series remainder bound ($e-{\sum }_{i=0}^n\frac{1}{i!}\le \frac{n+2}{(n+1)(n+1)!}$)

The positive tail after term $n$ in the reciprocal-factorial series for Euler's number satisfies $1/(n+1)!<e-{\sum }_{i=0}^{n}1/i!\le (n+2)/((n+1)(n+1)!)$. 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.