# Thread: alternating series remainder estimate

1. ## alternating series remainder estimate

Number 46

I found the first 7 terms to be .78343057
Since I want to 7 places I added the value when n=8 (5.96x10^-8) to my initial value
So .78343057<f(x)<.78343051

So I claim that the sum of the series should be .7834305

My teacher told me this is wrong but could not explain why. His value was .783431

Where is my mistake?

Listing from phone so hard to type out step by step at the moment.

2. ## Re: alternating series remainder estimate

You have rounded incorrectly. you have .78343057. Because 7 is closer to 10 than it is to 0, that is closer to .783431 than to .783430.

3. ## Re: alternating series remainder estimate

That doesn't make a lot of sense to me. I used my calculator to do the summation to n=7. I left out the other digits past 7 because that is the point where the two values differ. Even if it is a rounding error at 7, wouldn't it mean that instead of ...3057 I would have ...3060?

And besides, I thought the point of the exercise is to find the accuracy of the value to a certain decimal number. Why would I round it in the end? If he was just rounding I can see the mistake but it seems counter-productive if that is the case.

4. ## Re: alternating series remainder estimate

$|S - S_7| < \frac{1}{8^8}$

$-\frac{1}{8^8} < S - S_7 < \frac{1}{8^8}$

$S_7 - \frac{1}{8^8} < S < S_7 + \frac{1}{8^8}$

$0.7834305082 < S < 0.7834306274$