Approximate the sum of the series correct to four decimal places.
sum n=1,infinity ((-1)^(n-1) n^2)/(10^n)
I know how to finish the equation once I have the number I need to go up to, but I don't know how to find that number, can anyone help?

Question

Approximate the sum of the series correct to four decimal places.
sum n=1,infinity ((-1)^(n-1) n^2)/(10^n) 
I know how to finish the equation once I have the number I need to go up to, but I don't know how to find that number, can anyone help?

anonymous · Answer

You can do the same thing with this one, just solve another equality without the (-1)^(n-1)        n^2/(10^n)<0.0001  

This should give you the number of terms

anonymous · Answer

It doesn't give me an error though

anonymous · Answer

The error of the approximation?

anonymous · Answer

It wants me to find the sum.

anonymous · Answer

I think the easiest way to do this one is write out the terms, maybe the first 3 or 4.  and then find the first term that is less than 0.0001. 

Add all of the terms before this term and you get the approximate sum correct to 0.0001

anonymous · Answer

(cause i just realized in the equality was like 10^n which you can't really solve for easily )