# Looks like the Harmonic series

1. Oct 21, 2011

### Punkyc7

does let z$_{n}$=

$\frac{1}{n+1}$+$\frac{1}{n+2}$+$\frac{1}{n+3}$+.....+$\frac{1}{2n}$

does z$_{n}$ converge or diverge..

I want to say it diverges because it looks like the Harmonic series

2. Oct 21, 2011

### Punkyc7

Re: Sequences

yep I am positive

3. Oct 21, 2011

### Staff: Mentor

Re: Sequences

Each term in the sum above is >= 1/(2n), which is the smallest term, and there are n terms. This means that zn >= n/(2n) = 1/2.

Also, each term in the sum is <= 1/(n + 1), which is the largest term, and there are still n terms. This means that zn <= n/(n + 1).

One of the tests you have learned is applicable here.

Last edited: Oct 21, 2011
4. Oct 21, 2011

### Punkyc7

Re: Sequences

does that mean it converges?

5. Oct 21, 2011

### Staff: Mentor

Re: Sequences

Write down a few terms in your sequence.
z1 = 1/2
z2 = 1/3 + 1/4
z3 = 1/4 + 1/5 + 1/6

$$\lim_{n \to \infty} z_n~?$$

Note that I added some to my previous response.

6. Oct 21, 2011

### Dickfore

Re: Sequences

$$\begin{array}{rcl} z_{n + 1} - z_{n} & = & \left( \frac{1}{n + 2} + \ldots + \frac{1}{2 n} + \frac{1}{2 n + 1} + \frac{1}{2 n + 2} \right) \\ & - & \left(\frac{1}{n + 1} + \frac{1}{n + 2} + \ldots + \frac{1}{2 n} \right) \\ & = & \frac{1}{2 n + 1} + \frac{1}{2 n + 2} - \frac{1}{n + 1} \\ & = & \frac{1}{2 n + 1}- \frac{1}{2 n + 2} > 0 \end{array}$$

Post #4 holds the other clue.

Last edited: Oct 21, 2011
7. Oct 21, 2011

### Punkyc7

Re: Sequences

as n goes to infinity isnt the limit 1? because of zn <= n/(n + 1).

8. Oct 21, 2011

### Staff: Mentor

Re: Sequences

About all you can say is that 1/2 <= zn <= 1, but that doesn't mean that the limit has to be either of the endpoints.

9. Oct 21, 2011

### Punkyc7

Re: Sequences

since its bounded and increasing it converges doesnt

10. Oct 21, 2011

### Dickfore

Re: Sequences

no. look at, for example:
$$z_{2 n} = \left( \frac{1}{2n + 1} + \ldots + \frac{1}{3 n} \right) + \left(\frac{1}{3 n + 1} + \ldots + \frac{1}{4 n}\right) \le \frac{n}{2 n + 1} + \frac{n}{3 n + 1} = \frac{n (5 n + 2)}{(2 n + 1)(3 n + 1)}$$
$$z_{2 n} \le \frac{5 n^2 + 2 n}{6 n^2 + 5 n + 1} = \frac{1}{\frac{6 n^2 + 5 n + 1}{5 n^2 + 2 n}} = \frac{1}{\frac{6}{5} + \frac{\frac{13 n}{5} + 1}{n (5 n + 2)}} \le \frac{1}{\frac{6}{5}} = \frac{5}{6}$$

Thus, there is a subsequence that is definitely not within a neighborhood $\epsilon = 1/5$ of 1. So, 1 is not a limit of the sequence.

yes, but it's actually decreasing.

Last edited: Oct 21, 2011
11. Oct 21, 2011

### Punkyc7

Re: Sequences

ok since its monotonic decreasing and bounded it is convergent

12. Oct 21, 2011

### Dick

Re: Sequences

It is NOT decreasing, it's increasing. I'd suggest to Punkyc7 not to believe everything everyone tells you.

13. Oct 21, 2011

### Dickfore

Re: Sequences

yes, and the limit is somewhere between 1/2 and 5/6. Actually, the exact sum is expressible via elementary functions. Do you know what it is?

14. Oct 21, 2011

### Dickfore

Re: Sequences

15. Oct 21, 2011

### Dick

Re: Sequences

I did. And I have no idea what you are doing. z_1=1/2, z_2=7/12, z_3=37/60. Now you read post #6 and tell me what's wrong with it.

16. Oct 21, 2011

### Dickfore

Re: Sequences

there's an extra term -1/n that does not belong there.

17. Oct 21, 2011

### Dick

Re: Sequences

Perhaps. I didn't read it THAT carefully. But I do know z_n is an increasingly accurate estimate to log(2). And all of the z_n are underestimates. Think of it as like an integral estimate.

18. Oct 21, 2011

### Dickfore

Re: Sequences

I corrected it.

Look at my post #13.

19. Oct 21, 2011

### Dick

Re: Sequences

Ok, so you know what it is. That's good. I'd still suggest to Punkyc7 to think more about the problem. And maybe for you to maybe provide less details until Punkyc7 does that.

20. Oct 22, 2011

### HallsofIvy

Staff Emeritus
Re: Sequences

The original sum is a sum of positive terms. If we are always adding a positive number, the sum must be increasing.