sum of infinite series

[logistic_guy]

Estimate the sum of each convergent series to within $\displaystyle 0.01$.

(a) $\displaystyle \sum_{k = 0}^{\infty}(-1)^{k+1}\frac{2}{k!}$

(b) $\displaystyle \sum_{k=3}^{\infty}(-1)^{k}\frac{k}{2^{k}}$

 

[khansaheb]

Estimate the sum of each convergent series to within $\displaystyle 0.01$.

(a) $\displaystyle \sum_{k = 0}^{\infty}(-1)^{k+1}\frac{2}{k!}$

(b) $\displaystyle \sum_{k=3}^{\infty}(-1)^{k}\frac{k}{2^{k}}$

show us your effort/s to solve this problem.

 

[Steven G]

show us your effort/s to solve this problem.

logistic_guy doesn't do that.

 

[jonah2.0]

Beer drenched reaction follows.

logistic_guy doesn't do that.

I say nay.
He does do that.
But only after a demand or request is made.

 

[logistic_guy]

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.

$\displaystyle |S - S_9| = |-0.22 - (-0.22)| = 0 \leq 0.01$