Thread: Alternating Series Estimation Theorem

Use the alternating series estimation theorem to approximate the sum of $\displaystyle (-1)^n/(n!)$ from n=0 to infinity with an error of $\displaystyle .000005$.

I know that after summing a number of terms, the remainder(error) will be less than the first omitted term, so

I've set up the problem like this:

$\displaystyle .000005 < 1/(n+1)!$

Is this right, and if so, what's next?

Originally Posted by JewelsofHearts

Use the alternating series estimation theorem to approximate the sum of $\displaystyle (-1)^n/(n!)$ from n=0 to infinity with an error of $\displaystyle .000005$.

I know that after summing a number of terms, the remainder(error) will be less than the first omitted term, so

I've set up the problem like this:

$\displaystyle .000005 < 1/(n+1)!$

Is this right, and if so, what's next?

So you want the least integer $\displaystyle $$n$ such that:

$\displaystyle (n+1)!>20000$

To proceed you need either to use trial and error or some suitable approximation for the factorial (and then a numerical solution of the resulting equation).

(look up $\displaystyle $$7!$ and $\displaystyle $$8!$)

CB

Shouldn't it be (n+1)!<200000?

Originally Posted by JewelsofHearts

Shouldn't it be (n+1)!<200000?

Probably counting zeros is not something I am particularly careful about

Not only the zeros, but I think the direction of the inequality sign is supposed to be different. I really don't know, though. I don't understand the theorem.

Help, please!

error, $\displaystyle E < \dfrac{1}{(n+1)!} < 5 \times 10^{-6}$

$\displaystyle (n+1)! > \dfrac{1}{5 \times 10^{-6}} = 2 \times 10^5$

note the following factorial values ...

$\displaystyle 8! = 40320$

$\displaystyle 9! = 362880$

so, how many terms of the series would you need to add to get an error less than $\displaystyle 5 \times 10^{-6}$ ?

fyi, to check your work, note ...

$\displaystyle \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = \frac{1}{e}$