Approximating sum of alternating series with factorial terms

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
11 replies · 10K views
johnnyies
Messages
93
Reaction score
0

Homework Statement


[tex]\Sigma[/tex](-1)[tex]^{n+1}[/tex][tex]\frac{1}{n!}[/tex]

How many terms will suffice to get an approximation within 0.0005 of the actual sum? Find that approximation.

Homework Equations


No idea.

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.
 
Physics news on Phys.org
What does the alternating series theorem tell you?
 
as the magnitude of the terms are monotonically decreasing, and alternating, you could also look at the magnitude of a single term
 
Last edited:
. . . that doesn't even seem viable to me. . . it's essentially guessing until you get the right error?
 
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.
 
the magnitude of the error of n terms is less than the next n + 1 th term?
 
Right, so you're trying to solve

[tex]\frac{1}{(n+1)!} < 0.0005[/tex]

Find how big (n+1)! has to be and then what n would satisfy that.
 
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?
 
Yes, but n=6 also works, right?