# Divergence of 1/x

1. Jul 22, 2008

### hyper

http://img403.imageshack.us/img403/9478/roffelsw8.png [Broken] I really can't understand the last sentence, how do they get that the sum has to be smaller than k/2?

Last edited by a moderator: May 3, 2017
2. Jul 22, 2008

### HallsofIvy

Staff Emeritus
??? They don't. They get that it is larger than k/2.

They are saying that if, in of the sum
$$\sum_{n=1}^k \frac{1}{n}$$
you group the terms: 1/2, then the sum of the next 2 terms, then the sum of the next 4 terms, then the sum of the next 8 terms, etc., each time grouping 2k terms, each of those sums will be larger than 1/2 and so the sum of k such groups will be larger than k/2. Since that is unbounded, the sum itself does NOT converge.

3. Jul 22, 2008

### hyper

Sorry, meant to say larger. But how can you see that there is k groups?

4. Jul 22, 2008

### arildno

First off:
Try to be precise in your thinking, by answering to yourself (and us!):

What does the "k" index count?

5. Jul 22, 2008

### maze

Ahh I think I see the confusion. The term groupings increase in length, so that each grouping is twice as long as the previous one, but what you are interested in is not the length of the latest grouping of terms, but rather the total number of terms so far including all previous groupings. This comprehensive sum also doubles with each grouping since it adds up nicely as you can see:
length of latest grouping: 2 2 4 8 16 ...
total # of terms so far: 2 4 8 16 32 ...

The bound (sn > k/2) is also weak (I think?), a tighter bound you get if you actually work the sum would be sn > k/2 + 1. I think they just dropped the 1 without mentioning it, since it doesnt matter anyways.

Last edited: Jul 22, 2008
6. Jul 22, 2008

### HallsofIvy

Staff Emeritus
There are k groups by definition. All they are saying is that each group has sum larger than 1/2 so if there are k groups then the sum is greater than k/2.

7. Jul 22, 2008

### arildno

Note that each group FROM $2^{k}$ up to and including $2^{k+1}$has length:
$$2^{k+1}-2^{k}=2^{k}(2-1)=2^{k}$$

8. Jul 22, 2008

### hyper

K groups by definition?, where does one see that?

9. Jul 22, 2008

### arildno

What does "k" count?

10. Jul 22, 2008

### hyper

Sorry I don't quite understand your question. I mean n=2^k så k=ln(2)/ln(n).

But maybe you mean that for every k, we double n?

11. Jul 22, 2008

### n_bourbaki

Because they looked at it and evaluated what was going on. There really isn't anything anyone can say other than "look at it", since it is self evident. Look at the first term, the next two, the next four, the next 8, the next 16 and so on.

Last edited by a moderator: May 3, 2017
12. Jul 22, 2008

### arildno

I did NOT ask you how to express "k" in terms of "n"!
Again:
What type of quantities/objects does the "k" count, and for that matter, what does the "n" count?

Last edited: Jul 22, 2008
13. Jul 22, 2008

### hyper

I don't know.

14. Jul 22, 2008

### Redbelly98

Staff Emeritus
n counts a finite number of terms in the series.
k counts the number of groupings of those terms, ignoring the 1st term "1".
n = 2^k is the number of terms included in the first k groupings of terms.

For example:
If k=3 then n = 2^3 = 8.
In the first 8 terms (i.e. up to the term "1/8"), there are k=3 groupings not counting the "1" term:
"1" is not considered part of any group
"1/2" is gouped by itself
"1/3 + 1/4" is the 2nd grouping of terms
"1/5 + ... + 1/8" is the 3rd grouping

That's 8 terms and 3 groups of terms. The number of terms is eight, and three is the number of groupings.

15. Jul 23, 2008

### arildno

As you can see, your primary problem has been that you simply didn't know what the original text talked about.
Read Redbelly's answer carefully, and see if you understand the orinal problem better now.