# Alternating Series Theorem (remainder formula)

• Jul 22nd 2009, 02:36 PM
Hikari
Alternating Series Theorem (remainder formula)
The Alternating Series Estimation theorem states:

If s= $[summation](-1)^n*b_n$ from n=1 to infin. is the sum of a convergent, alternating series that satisfies:
a) 0 < b_n+1 < b_n
b) lim b_n -> 0 as n-> infin

then:

|R_n|= s - s_n <= b_n+1

So the problem I'm given is this:

How many terms of the series do I need to add in order to find the sum to the indicated accuracy?

$[summation](-2)^n/n!$; error < 0.01

Okay, so I can use the theorem mentioned above because this series is both alternating and convergent (the positive part of the sum is decreasing for all n, and the lim 1/n! -> 0 as n-> infin, satisfying both conditions for convergence).

So... |R_n| <= b_n+1 < 1/100... right?

b_n+1 = 1/(n+1)! < 1/100

So how do I solve for n? Looking at the solution for this problem, it seems like the person kept guessing random values of n until (-2)^n/n! was less than 1/100, which I don't really understand. The answer ended up being 7. That is, the sum of 7 terms to keep within the given accuracy.

I guess I don't know how to do these kinds of problems when the "negative part of the series" is something other than (-1)^n.
• Jul 22nd 2009, 03:03 PM
Calculus26
first of all bn = 2^n/n! since (-2)^n = (-1)^n 2^n

you want 2^(n+1)/[(n+1)!] < .01

There is no analytic way of solving this--simply make a table of values

using your calculator and look for the smallest n

for which 2^(n+1)/[(n+1)!] < .01

Or if you prefer crank it aout by hand try n= 2 if that doesn't work try n = 3 etc.
• Jul 22nd 2009, 03:21 PM
Hikari
Eureka! Thanks a bunch!
• Jul 23rd 2009, 07:36 AM
calc101
Just re-writing with latex:

If $s = \sum_{n=0}^{\infty} (-1)^n b_n$ is the sum of a convergent, alternating series that satisfies:

a) $0 < b_n+1 < b_n$

b) $\lim_{n=\infty} b_n = 0$

then:

$|R_n|= s - s_n <= b_n+1$

Problem: How many terms of the series do I need to add in order to find the sum to the indicated accuracy?

$\sum_{n=0}^{\infty} \frac{(-2)^n}{n!}$; error <0.01