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

Question

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

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?

Answer 1

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

References:

http://ecademy.agnesscott.edu/~lriddle/apcalculus/approxSeries.pdf

Answer 2

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.