# Sum of alternating series

1. Mar 13, 2005

### shan

We were given the series:
1/2^6 - 1/2^8 + 1/2^10 - 1/2^12 + ...

And asked to find the general term, an, which I worked out to be (-1)^(n+1)/2^(4+2*n).
To see if it converged, I used the alternating series test and found that it converged ie the lim as n tends to infinity = 0 and the absolute values of the terms are decreasing.

The last part of the question asks for the sum of the series but I don't know how to find it. I searched on the internet but only found the alternating series estimate theorem which we haven't been taught and don't know how to use

2. Mar 13, 2005

### Data

I'll give you a few little hints. First, you're looking for solutions that would make the problem much more complicated than it is.

Secondly, is the series absolutely convergent? Can this help you at all?

3. Mar 13, 2005

### physicsCU

Since you are asked to sum the series, you can do it either using geometric series or telescoping.

Geometric is easier. Calculate the first two terms of the series. The first term is a. The second term is a*r.

To solve, plug in a, solve for r.

The sum is a/(1-r). Note: r must be less than zero.

4. Mar 13, 2005

### Data

Her series is alternating, so she can't do that directly. Also, in order for that to work, $$r < 0$$ is not the right condition. I think you meant $$|r| < 1$$ :)

Last edited: Mar 13, 2005
5. Mar 13, 2005

### Data

Actually she can do it that way, if she chooses $$r$$ right! Good observation~

Last edited: Mar 13, 2005
6. Mar 13, 2005

### shan

hmmm I know it's conditionally convergent... But I don't see how that helps

I did consider using geometric but I don't know ... does it still work if the equation is not in the form of r^n? Or is that what you mean, by choosing the right r?

When I tried using the sum to infinity for geometric, I used a= 1/2^6 and r= -1/2^4 = -1/4 and got 0.0125. Is that the right r?

And by the way... I'm a girl

Last edited: Mar 13, 2005
7. Mar 13, 2005

### Data

I think that if you check, you'll find it's actually absolutely convergent.

Conditionally convergent series converge to a different value for each ordering of the terms of the series, so if it were conditionally convergent, you'd have to evaluate the sum exactly the way it's written (which would be very difficult).

Since it's absolutely convergent, you can reorder the terms however you like, or, indeed, split the series into two different series by choosing certain terms to put in each. Do you see a way that you can do this that will make it easier to evaluate?

And yes, in this case you can actually use the geometric series formula, although the choice of r isn't as obvious as it usually is!

sorry! Corrected :)

8. Mar 13, 2005

### shan

Oh yes I see now :)

So... create one series with the positive numbers and the other series with the negative numbers?

9. Mar 13, 2005

### Data

That's what I did!

The geometric series way is easier though, if you want to use that. The choice for r that works is $$\left(-\frac{1}{4}\right)$$, and the series is equal to

$$-\sum_{k=3}^\infty \left( -\frac{1}{4}\right) ^k$$

There you can just apply the geometric series formula and subtract away the missing k=1 and k=2 terms.

Using the method in which you split the series up, you have to apply the geometric series formula as well anyways, so it takes longer.

10. Mar 13, 2005

### shan

Sorry I'm getting confused again lol

Using the geometric formula that you put up: $$-\sum_{k=3}^\infty \left( -\frac{1}{4}\right) ^k$$
the sum of that was -0.8
If I subtract away the missing k=1 and k=2 terms...
-0.8 - 0.0625 + 0.25 = -0.6125

I also tried the other way, making two other series with r=1/16 and a=1/2^6 for the positive values and r=1/16 and a=-1/2^8 for the negative values (the sums being 8/195 and -4/105 respectively) and added them together to get 4/1365....

Now I'm sure I've done something wrong but I'm not sure what and to which one

11. Mar 13, 2005

### shan

Wait, never mind, I must've put in the wrong numbers on my calculator.... I got it, thanks ^^ It's 0.0125 :)

12. Mar 13, 2005

### Data

Since you have the method now, I don't mind showing you my solutions. Hopefully they'll help you figure out where the mistake is (maybe I made one!).

Using the split strategy:

$$\frac{1}{2^6} - \frac{1}{2^8} + \frac{1}{2^{10}} - . \ . \ . = \sum_{k=2}^\infty \left(\frac{1}{2}\right)^{4k-2} - \sum_{k=2}^\infty \left(\frac{1}{2}\right)^{4k}$$

$$= 4\sum_{k=2}^\infty \left(\frac{1}{2}\right)^{4k} - \sum_{k=2}^\infty \left(\frac{1}{2}\right)^{4k} = 3\sum_{k=2}^\infty \left(\frac{1}{16}\right)^{k}$$

$$= 3\left( \frac{\frac{1}{16}}{1-\frac{1}{16}} - \frac{1}{16} \right)= 3\left(\frac{1}{15} - \frac{1}{16}\right) = 3\frac{1}{240} = \frac{1}{80}$$

And now using the direct geometric series method:

$$\frac{1}{2^6} - \frac{1}{2^8} + \frac{1}{2^{10}} - . \ . \ . = -\sum_{k=3}^\infty \left(-\frac{1}{4}\right)^k$$

$$= - \left( \frac{-\frac{1}{4}}{1 + \frac{1}{4}} + \frac{1}{4} - \frac{1}{16}\right) = - \left( -\frac{1}{5} + \frac{1}{4} - \frac{1}{16}\right)$$

$$= -\left( \frac{1}{20} - \frac{1}{16}\right) = -\left( -\frac{4}{320}\right) = \frac{1}{80}$$

:)

Last edited: Mar 13, 2005
13. Mar 14, 2005

### shan

Thanks very much for that Data :) Your working out is so much neater than mines lol

14. Mar 14, 2005

### Data

I write 20 page physics lab reports in TeX every week, so I get very used to the math syntax :)

15. Mar 14, 2005

### HallsofIvy

Staff Emeritus
Frankly, I don't see any reason to separate this into two sums (or worry about whether it is absolutely convergent or not).
This is, as physicsCU said at the beginning, a geometric series.

The first term is $\frac{1}{2^6}$ which we can take as "a" and the second term is $\frac{-1}{2^8}= \frac{1}{2^6}\frac{-1}{4}$ so r is -1/4. You can check to see that all other terms are just the previous term times -1/4 so this is in fact, a geometric series.
$$\sum_{k=0}^{\infty}\frac{1}{2^6}\left(\frac{-1}{4}\right)^k= \frac{1}{64}\frac{1}{1+\frac{1}{4}}= \frac{1}{80}$$

Last edited: Mar 14, 2005
16. Mar 14, 2005

### Data

Yes. Like I said, using a plain geometric series is easier~