Estimating a sum of an Infinite series

[G01]

How many terms of :

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}

do you have to add to get an error < .01

Alright, I used the Alternating Series Estimation Theorem since the terms are decreasing and the terms approach 0.

So, by the theorem, .01 < = b_{n+1} so

1/(n+1)^2 = 1/100
(n+1)^2 = 100
n+1 = 10

So this means that in order to get this error, we have to add 9 terms right? The back of my book says 10 is the answer. Why is that?

 

[0rthodontist]

You have a less than or equal sign in one place and a less than sign in the other place. You've shown that the error is less than or equal to 1/100 if you evaluate to 9 terms, but you need one more to show that it's actually less than by this method.

 

[G01]

AH HAH! Thats it!
Its supposed to be a less than sign, thanks!

 

[dextercioby]

BTW, the sum is exactly \frac{\pi^{2}}{12}.

Daniel.