Simple Sum

1. Feb 13, 2006

cscott

How can I find the sum of $$1/2 - 1/4 + 1/8 - 1/16 + ... - 1/256$$?

Do I need to group the positive and negative terms?

2. Feb 13, 2006

TD

Do you mean the finite sum, as you wrote it, or an infinite sum (the series)?

3. Feb 13, 2006

cscott

Finite sum.

4. Feb 13, 2006

TD

In that case: yes you could add the positive terms and substract the sum of the (absolute values) of terms with a negative sign. It would be rather annoying (bu still doable) work to put them all on the same denominator, can you use a calculator or not?

5. Feb 13, 2006

cscott

No calculator allowed :\

I only know sigma notation and the formula's for the partial sums of geometric and arithmetic series'.

6. Feb 13, 2006

TD

Oh of course, but that's fine.

The sum as a whole can be written as:

$$\sum\limits_{n = 1}^8 {\frac{{\left( { - 1} \right)^{n + 1} }}{{2^n }}}$$

Now, can you split it in two sums and find the general formula for both?
Or even if you can't find the sigma-notation, can you try to become two geometric series if you look at the positive terms and at the negative terms seperately?

7. Feb 13, 2006

cscott

Oops, I had that answer but I made the mistake of thinking there was 256 terms :\

How do I figure out the common ratio if it varies from positive to negative?

I will try and split it up into two sums in the mean time...

8. Feb 13, 2006

TD

Well, if you can find the common ratio by dividing a term by its precessor (is that an English word?) Anyway, if the sequence with terms t_n is geometric, than t_(n+1)/t_n = r with r constant for all n. You can check this and find r this way.

9. Feb 13, 2006

cscott

I get:

$$\sum\limits_{n = 1}^4 {\frac{1}{2 \cdot 4^{n - 1}}} - \sum\limits_{n = 1}^4 {\frac{1}{4 \cdot 4^{n - 1}}}$$

10. Feb 13, 2006

TD

Looks good, but have you checked whether the initial problem wasn't geometric already?

11. Feb 13, 2006

cscott

r = -1/2

So,

$$\sum\limits_{n = 1}^8 {\left[\frac{1}{2} \cdot \left (-\frac{1}{2}\right)^{n-1}\right]}$$

Last edited: Feb 13, 2006
12. Feb 13, 2006

TD

Looks good again

13. Feb 13, 2006

cscott

14. Feb 14, 2006

cscott

I have another problem:

$$S = 1/2 - 1/3 + 1/4 - 1/5 + ...$$

Find $S_{100}$

So, $$S_n = \frac{n}{2}(t_1 + t_n) = \frac{100}{2}\left({\frac{1}{2} - \frac{1}{101}\right)$$

This gives me 24.505 when I think it should be 0.301927

Also, how can I associate the terms to show that 1 > S > 0? Does grouping positive terms and subtracting the negative terms show this?

15. Feb 14, 2006

TD

The formula you used to find the n-th partial sum is the one for arithmetic series, is that the case here? If you think so, what is the common difference then?

16. Feb 14, 2006

cscott

I guess there is no common difference... but I see no common ratio either?

Last edited: Feb 14, 2006
17. Feb 14, 2006

TD

Correct, it's not arithmetic nor geometric...

18. Feb 14, 2006

cscott

I see it's a harmonic series, so the recipricals are arithmetic. Does that help me?

19. Feb 14, 2006

TD

I'm not sure, are you supposed to determine this partial sum by some calculation again (no calculator)?

20. Feb 14, 2006

cscott

Yes... never calculator :p

I only ever got a forumla for the partial sum of a geometric and arithmetic series.