# Alternating series, error estimation & approximation

1. Apr 2, 2010

### johnnyies

1. The problem statement, all variables and given/known data
$$\Sigma$$(-1)$$^{n+1}$$$$\frac{1}{n!}$$

How many terms will suffice to get an approximation within 0.0005 of the actual sum? Find that approximation.
2. Relevant equations
No idea.

3. The attempt at a solution
What I tried doing is setting my absolute value of the series less than 0.005, but I have no idea how to get rid of that factorial.

2. Apr 2, 2010

### vela

Staff Emeritus
What does the alternating series theorem tell you?

3. Apr 2, 2010

### lanedance

as the magnitude of the terms are monotonically decreasing, and alternating, you could also look at the magnitude of a single term

Last edited: Apr 2, 2010
4. Apr 2, 2010

### johnnyies

so I just plug in numbers?

5. Apr 2, 2010

### vela

Staff Emeritus
Yeah.

6. Apr 2, 2010

### johnnyies

. . . that doesn't even seem viable to me. . . it's essentially guessing until you get the right error?

7. Apr 2, 2010

### vela

Staff Emeritus
Well, you shouldn't be making a wild guess. What exactly are you trying to do? You never answered my question about what you know about alternating series, in particular, about the error.

8. Apr 3, 2010

### johnnyies

the magnitude of the error of n terms is less than the next n + 1 th term?

9. Apr 3, 2010

### vela

Staff Emeritus
Right, so you're trying to solve

$$\frac{1}{(n+1)!} < 0.0005$$

Find how big (n+1)! has to be and then what n would satisfy that.

10. Apr 3, 2010

### johnnyies

invert both sides with inequality switched

(n+1)! > 2000

so (6+1)! = 5040 > 2000

so all n > 6 will make an error less than 0.0005?

11. Apr 3, 2010

### vela

Staff Emeritus
Yes, but n=6 also works, right?

12. Apr 3, 2010

### johnnyies

ya, much thanks!