# 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. Apr 18, 2015 at 8:36
• oh wait let me fix that thanks
– Elsa
Apr 18, 2015 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, 2015 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, 2015 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, 2015 at 8:47

You could use the Calabrese criteria (see the references) to approximate the sum of alternating series:

References:

This is horribly wrong proposition but that was the case in 1955 at least in some textbooks.

Using the alternating series error estimation theorem, what is the smallest number of terms

It cannot provide smallest number in all cases! Only some number where error is at least as specified but in reality error can be smaller.

$$0.0000005\le\frac{(N+1)}{(N+1)^2+8}$$

gives $$N \le -0.9999954999657348$$ and $$2000000.9999954998\le N$$

So the answer is 2 million terms, now we need to verify there is no smaller value, like 1 million (which is 2 times less, that is how Calabrese criteria error bond works).

Well, you can check this with Wolfram Mathematica first.

The sum of the series can be found in symbolic form: $$\frac{1}{4} \left(\psi ^{(0)}\left(i \sqrt{2}+1\right)-\psi ^{(0)}\left(i \sqrt{2}+\frac{1}{2}\right)+\psi ^{(0)}\left(1-i \sqrt{2}\right)-\psi ^{(0)}\left(\frac{1}{2}-i \sqrt{2}\right)\right)$$

where $$\psi ^{(0)}$$ is polygamma function.

We can see that it is actually 1 million terms, not 2 million.

+1 to the guy above who gave the citation.