Alternating series, error estimation & approximation

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The discussion focuses on estimating the error of the alternating series Σ(-1)^{n+1}/n! to determine how many terms are needed for an approximation within 0.0005 of the actual sum. Participants clarify that the error of an alternating series can be estimated using the next term in the series, leading to the inequality 1/(n+1)! < 0.0005. By solving this inequality, it is established that n must be greater than or equal to 6 to ensure the error is within the desired range. The conclusion confirms that using 6 terms will suffice for the approximation. This approach effectively utilizes the properties of alternating series to achieve the desired accuracy.
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Homework Statement


\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.

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.
 
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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:
so I just plug in numbers?
 
Yeah.
 
. . . 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

\frac{1}{(n+1)!} &lt; 0.0005

Find how big (n+1)! has to be and then what n would satisfy that.
 
  • #10
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
Yes, but n=6 also works, right?
 
  • #12
ya, much thanks!
 

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