Alternating series error estimatation theorem for $\sum_1^{\infty} (-1)^{n+3}\frac{n}{n^2+8}$

$$\sum_1^{\infty} (-1)^{n+3}\frac{n}{n^2+8}$$

Here is my question: Using the alternating series error estimation theorem, what is the smallest number of terms needed to estimate the entire sum with an error of magnitude less than $10^{-6}$?

So from what I know about the alternating series error estimation theorem, it's when you have to list out the terms and adding up a certain number of terms makes the error of the estimation the next term that was left out of the group of terms.

So I wrote out the first few terms, is that how you approach this problem? Am I supposed to find the term with the 0.000001 place?

Here are the first few terms:

0.11 - 0.166 + 0.176 - 0.16 + 0.151

So far I do not see any term to that place. Is there some other way to do this problem?

• I suppose that you did not finish your last sentence. – Claude Leibovici Apr 18 '15 at 8:36
• oh wait let me fix that thanks – Elsa Apr 18 '15 at 8:37
• So your problem is that none of the five first terms is less than 10^{-6}? Sorry but (1) does this really come as a surprise? and (2) sure you have no idea how to overcome this obstacle? – Did Apr 18 '15 at 8:44
• @Did well um ya I just started learning about this theorem, so (1)ya it comes as a surprise and (2)my only idea is to keep listing out terms. Is that how it goes? – Elsa Apr 18 '15 at 8:46
• Imagine that the first suitable term is at n = 10^6... Still think this is how it goes? And if you were asked for some rest less than 10^{-12}... – Did Apr 18 '15 at 8:47