# Alternating series

1. May 1, 2005

$$\sum_{n=1}^{\infty} a_n = 1 - \frac {(0.3)^2}{2!} + \frac {(0.3)^4}{4!} - \frac {(0.3)^6}{6!} + \frac {(0.3)^8}{8!} - ...$$

how many terms do you have to go for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series?

the answer to this question is 4, but i dont know how the book got 4. Probably a real easy question, but im really confuse since there are no examples i can find, so can someone help? i really dont even know where to start, but i found this:

$$|s-s_n| \leq |s_n+1 - s_n| = b_n +1$$

any help will be appreciated

2. May 1, 2005

### OlderDan

Calculate the values of each of the terms and note the progression of the sizes of them. Isn't that inequality supposed to be

$$|s-s_n| \leq |s_{n+1} - s_n| = b_{n +1}$$

It is saying that the absolute value of the remainder after n terms will be no greater than the absolute value of the difference between the sum to n + 1 terms and the sum to n terms. Another way of saying that is look at the next term.

http://www.mathwords.com/a/alternating_series_remainder.htm

For an alternating series, you only have to look at the magnitude of the first term you are dropping from the sum to estimate the remainder.

3. May 1, 2005

well the 4th term is $$\frac{(0.3)^6}/{6!}$$ but it comes out to .000001

and the 5th term has 8 zeros, so the 4th term is closer to the value 0.0000001. so is that how the book got 4th term as an answer?

4. May 1, 2005

### whozum

The question is asking to find $n$ such that:

$$s - s_n < 10^{-7}$$

Once youve found the first n, theres no need to go further.

5. May 1, 2005

### OlderDan

It is not a question of being closer. It is a question of greater than or lesser than. The fourth term is ten times bigger than the permitted remainder, so you have to keep it. You would have to keep it even if its value were .00000010000. . .000001. The first term you can leave out is the first term that is smaller than .00000001. That is the fifth term. That is how the book got the answer.