Estimating the Value of a Series — Question 4

*Question 4 Approximate $\displaystyle e^{-1}=\sum_{n=0}^{\infty}(-1)^n/n!$ using $S_N=\sum_{n=0}^{N}(-1)^n/n!$. Find the least final index $N$ that guarantees correct rounding through six decimal places.

*Question 4 – Solution

The magnitudes $1/n!$ decrease to $0$, so $$|R_N|\le\frac1{(N+1)!}.$$ The tolerance $|R_N|<5\times10^{-7}$ suggests a candidate; correct rounding also requires a bracket within a single rounding interval.

$$9!=362{,}880,\qquad 10!=3{,}628{,}800.$$ Hence $$\frac1{9!}\approx2.76\times10^{-6}>5\times10^{-7},\qquad \frac1{10!}\approx2.76\times10^{-7}<5\times10^{-7}.$$ Therefore $N+1=10$, so $\boxed{N=9}$.

The alternating sign gives $$0.3678791887<S<S_9+1/10!<0.3678794643.$$ Both endpoints lie in $[0.3678785,0.3678795)$ and round to $0.367879$. Also $S_8\approx0.3678819444$ and $S_7\approx0.3678571429$; the earlier even and odd partial sums lie still farther away. Hence no smaller final index gives this rounded value.

Ten terms (indices $0$ through $9$) suffice. They give $S_9\approx0.3678791887$, which rounds to $0.367879$, the same six-decimal value as $e^{-1}$.