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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?