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Choose the reciprocal-factorial summation eb for asymptotically high accuracy, counting multiplications and divisions as arithmetic work. For tolerance $ϵ$, linear powering in part (a)(i) takes $O({ϵ}^{-1})$ multiplications; binary exponentiation in part (a)(ii) reduces this to $O(\log (1/ϵ))$. Part (c) also has $O(\log (1/ϵ))$ arithmetic work, since its required $n$ grows like ${ϵ}^{-1/2}$, though its constants and cancellation error differ.

By the reciprocal-factorial series remainder bound, the omitted positive tail satisfiesThus the factorial denominator grows much faster than an exponential in the number of iterations. For $d$ decimal places, setting $ϵ$ of order $10^{-d}$ gives $n=\Theta (d/\log d)$ by Stirling's approximation, rather than $\Theta (d)$ binary-power arithmetic steps. This is a comparison in an exact-arithmetic operation-count model; the stated Big O notation does not determine the exact winner for every small tolerance. Taken literally, eb uses zero multiplications and only divisions. Fixed double-precision floating-point format cannot deliver arbitrarily many correct decimal places, and the precision-dependent cost of arbitrary-precision arithmetic is a separate issue.

Use the reciprocal-factorial sum for increasing accuracy. Let $d$ be the requested number of decimal places and use absolute error tolerance proportional to $10^{-d}$. Expanding $\log (1+1/n)$ gives $e_a(n)=e-e/(2n)+O(n^{-2})$ and $e_c(n)=e+e/(24n^2)+O(n^{-3})$, so these are genuine first nonzero error orders. The first approximation needs $n$ of order $10^d$. Even its fast implementation therefore uses $\Theta (d)$ multiplications, while the linear version uses exponentially many in $d$. The third approximation needs $n$ of order $10^{d/2}$, but its fast powering still uses $\Theta (d)$ multiplications, with two parallel accumulators and a cancellation-prone subtraction.

For the Taylor series sum, Stirling's formula gives $\log (n!)=\Theta (n\log n)$, so a sufficient term count grows as $n=\Theta (d/\log d)$. Its divisions therefore grow more slowly than the multiplication counts of either fast-power alternative. More concretely, its positive tail satisfiesThe first omitted term also gives a lower bound $1/(n+1)!$. This is the reciprocal-factorial series remainder bound, supplying an explicit stopping certificate rather than relying only on Big O notation. It follows by bounding every successive tail denominator factor below by $n+2$.

The sum also avoids subtraction of nearly equal large values and has small positive terms, making it numerically preferable. For a particular small accuracy request, constants and actual term counts may be compared directly; an asymptotic error bound alone does not prove a universal finite-input operation ranking. With ordinary Double, sufficiently many tiny positive terms stop changing the rounded sum, so beyond machine precision a higher-precision version is necessary regardless of the formula.