Theorem.
\(\displaystyle |S - S_n| \leq a_{n+1}\)
where \(\displaystyle |S - S_n|\) is the absolute error.
For \(\displaystyle \bold{(a)}\), we are given:
\(\displaystyle |S - S_n| \leq 0.01\)
Then,
\(\displaystyle |S - S_n| \leq a_{n + 1} = \frac{2}{(n + 1)!}\)
\(\displaystyle \frac{2}{(n + 1)!} \leq 0.01\)
This gives \(\displaystyle n \geq \lceil4.29487\rceil = 5\).
So, summing \(\displaystyle 5\) terms will guarantee an error no more than \(\displaystyle 0.01\).
\(\displaystyle \sum_{k = 0}^{\infty}(-1)^{k+1}\frac{2}{k!} = S \approx S_5 = \sum_{k = 0}^{4}(-1)^{k+1}\frac{2}{k!} = -0.75\)
Actual sum:
\(\displaystyle \sum_{k = 0}^{\infty}(-1)^{k+1}\frac{2}{k!} \approx -0.74\)
Checking the absolute error.
\(\displaystyle |S - S_5| = |-0.74 - (-0.75)| = 0.01 \leq 0.01\)
For \(\displaystyle \bold{(b)}\):
\(\displaystyle |S - S_n| \leq a_{n + 1} = \frac{n + 1}{2^{n + 1}}\)
\(\displaystyle \frac{n + 1}{2^{n + 1}} \leq 0.01\)
This gives \(\displaystyle n \geq \lceil 8.960002\rceil = 9\).
So, summing \(\displaystyle 9\) terms will guarantee an error no more than \(\displaystyle 0.01\).
\(\displaystyle \sum_{k = 3}^{\infty}(-1)^{k}\frac{k}{2^k} = S \approx S_9 = \sum_{k = 3}^{11}(-1)^{k}\frac{k}{2^k} \approx -0.22\)
Actual sum:
\(\displaystyle \sum_{k = 3}^{\infty}(-1)^{k}\frac{k}{2^k} \approx -0.22\)
Checking the absolute error.
\(\displaystyle |S - S_9| = |-0.22 - (-0.22)| = 0 \leq 0.01\)
