- #1
danni7070
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Homework Statement
Find the smallest integer n that ensures that the partial sum [tex] s_n [/tex] approximates the sum s of the series with error less than 0.001 in absoulute value.
[tex] \sum_{n=1}^\infty (-1)^{(n+1)}\frac{n}{2^n} [/tex]
Homework Equations
[tex] {\abs{s-s_n}} <= s_{(n+1)} = a_{(n+1)}[/tex]
(How can I do absolute value in LaTex?)
The Attempt at a Solution
I've got this far.
error <= first term = [tex] \frac{n+1}{2^{n+1}} [/tex]
(because (-1)^n+1 doesn't matter in this example)
[tex] s - s_n < \frac{n+1}{2^n+1} < 0.001 [/tex]
Here I get lost.
From example in my textbook which says
[tex] a_n = \frac{1}{1+2^{n+1}} [/tex]
This error is less than 0.001 if [tex] 1+2^{(n+1)} > 0.001 [/tex] Since [tex] 2^{10} = 1024, n+1 = 10 [/tex] will do. Hence n = 9.
In my example I have n on both sides of the fraction. So what do I do?