Simple Sum with positive and negative terms

[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?

 

[TD]

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

 

[cscott]

Finite sum.

 

[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?

 

[cscott]

No calculator allowed :\

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

 

[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?

 

[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...

 

[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.

 

[cscott]

I get:

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

 

[TD]

I get:

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

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

 

[cscott]

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

r = -1/2

So,

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

 

[TD]

r = -1/2

So,

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

Looks good again

 

[cscott]

Looks good again

Thanks for your help!

 

[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?

 

[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?

 

[cscott]

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

 

[TD]

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

 

[cscott]

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

 

[TD]

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

 

[cscott]

Yes... never calculator :p

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