# Homework Help: Proving this series converges or diverges

1. Jul 6, 2013

### bonfire09

1. The problem statement, all variables and given/known data
Prove $y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ where $n\in\mathbb{N}$ is convergent or divergent.

2. Relevant equations

3. The attempt at a solution
I am stuck on this. I think its divergent but I am having trouble proving that it is divergent. I tried to break up the sequence using partial sums but I don't see a pattern. What I also don't get is what $y_1$ is equal too. Is it $y_1=\frac{1}{1+1}+\frac{1}{1+2}+...+\frac{1}{2*1}$ or is it equal to $y_1=\frac{1}{1+1}$? I've been stuck for quite a while.

Last edited: Jul 6, 2013
2. Jul 6, 2013

### Mandelbroth

From what I understand, it looks like this:
$$y_n=\sum_{k=n+1}^{2n}\frac{1}{k}.$$
That help?

3. Jul 6, 2013

### bonfire09

Yes but how would I go about proving it is divergent. I was thinking of using $\frac{1}{n^2}+...+\frac{1}{n^2}$ since $\frac{1}{n+1}+...+\frac{1}{2n}>\frac{1}{n^2}+...+\frac{1}{n^2}$?{Edit*I don't think it works}. I was thinking of using $\frac{1}{n+1}+...+\frac{1}{n+1}$. I don't think the way you wrote the series works. I think the limits have to be adjusted.

Last edited: Jul 6, 2013
4. Jul 6, 2013

### HallsofIvy

No, that doesn't work. Since the denominators are all "of order n", it would be better to compare the terms to 1/n (or your suggestion 1/(n+1)). For all i between 1 and n, $1/(n+i)\ge 1/(2n)$

5. Jul 6, 2013

### Mandelbroth

I...don't think so. Check again. I'm pretty sure my expression for the summation is correct.

I'd use a test of convergence. Are you familiar with any?

6. Jul 6, 2013

### bonfire09

Well this what I get so far that $\sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1}$ Hence its divergent. I think it works since $\sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}$ ≤ $\sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$

Last edited: Jul 6, 2013
7. Jul 6, 2013

### lurflurf

This is a little bit subtle

$$\frac{1}{2}=\frac{n}{2n}<\sum_{k=1}^n \frac{1}{n+k}<\frac{n}{n+1}<1$$

hint:think log(n)

hint2:
if

$$H_n=\sum_{k=1}^n \frac{1}{k}$$

then

$$\sum_{k=1}^n \frac{1}{n+k}=H_{2n}-H_{n}$$

Last edited: Jul 6, 2013
8. Jul 6, 2013

### bonfire09

so is the way I have it not correct then?

9. Jul 6, 2013

### lurflurf

^The way you wrote the sum is right, but you cannot conclude it diverges from what you have done.

10. Jul 6, 2013

### Dick

If you know what a Riemann sum is, write one for the integral of the function 1/x from x=1 to x=2.

11. Jul 6, 2013

### bonfire09

Ok I will try other methods. I would think there would be a much simpler way of doing this because Bartle (book im using) presents convergence and divergence of series before he introduces integration and differentiation.

12. Jul 6, 2013

### Dick

Fair enough. You can also show a sequence converges if you can show it's monotone and bounded. Can you show $y_{n+1}-y_{n}$ is positive so it's monotone increasing and, as lurflurf indicated, bounded above?

13. Jul 6, 2013

### bonfire09

So $y_{n+1}= \frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n+1}$ and $y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$. So $y_{n+1}-y_n= \frac{-1}{n+1}+\frac{1}{2n+1}$. I think that is how it supposed to go.

14. Jul 6, 2013

### Dick

Right idea, but the last term in $y_{n+1}$ is $\frac{1}{2(n+1)}$, not $\frac{1}{2n+1}$. Do it a little more carefully. It might be good to write out the n=1, n=2, n=3 cases explicitly, then do some differences and compare with what you think the answer is symbolically. That's what I do.

Last edited: Jul 6, 2013
15. Jul 6, 2013

### bonfire09

Oh yeah I forgot. $y_{n+1}= \frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n+5}$ and $y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ So $y_{n+1}-y_n=\frac{-1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}$ which seems to be decreasing that is $y_{n+1}\leq y_n$

16. Jul 6, 2013

### Dick

Ok, so why do you think $\frac{-1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}$ is negative? I don't. Just realized 2n+5 has got to be a typo, since your difference is correct.

Last edited: Jul 6, 2013
17. Jul 6, 2013

### bonfire09

Oh yes its increasing. Now once I show that $\frac{-1}{n+1} \leq \frac{1}{2n+1}+\frac{1}{2n+2}$. Now I think I'm going to have to use a epsilon proof to show $y_n$ its convergent which I need some help with.

18. Jul 6, 2013

### Dick

That's a little garbled too. My whole point is that there is likely a theorem in Bartle that if a sequence is monotone increasing and bounded above, then it converges. Use the theorem. No need for epsilons if you've got the theorem.

Last edited: Jul 6, 2013
19. Jul 6, 2013

### bonfire09

It says 3.3.2 Monotone Convergence Theorem A monotone, sequence of real numbers is
convergent if and only if it is bounded. Further: (a) If X = (xn) is a bounded increasing sequence, then $lim(x_n) = sup\{x_n : n ε N\}$. I some how have to show $y_n$ is bounded and all I know its increasing so it is monotone.

20. Jul 6, 2013

### Dick

lurflurf already hinted you that it's bounded. You are summing n numbers and they are all less than 1/n, so? Doesn't that mean it's bounded? Are you clear on the monotone bit? Why is $\frac{-1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}$ positive?

21. Jul 6, 2013

### bonfire09

Well $\frac{-1}{n+1}<\frac{1}{2n+1}+\frac{1}{2n+2}$ since if we start from $\frac{-3}{n+1}<\frac{2}{2n+1}=\frac{-2}{n+1}<\frac{2}{2n+1}+\frac{1}{n+1}=\frac{-2}{n+1}<\frac{2}{2n+1}+\frac{2}{2n+2}=\frac{-1}{n+1}<\frac{1}{2n+1}+\frac{1}{2n+2}$ Thus its positive. Oh yeah I saw the hint. Since its bounded and monotone its convergent by the monotone convergent theorem.

22. Jul 6, 2013

### Dick

$\frac{-1}{n+1}<\frac{1}{2n+1}+\frac{1}{2n+2}$ is certainly true. But that's useless. You want to prove $\frac{-1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2} \gt 0$. That's completely different. Do the algebra and combine the fractions.

23. Jul 6, 2013

### bonfire09

Well $\frac{1}{n+1}<\frac{1}{2n+1}+\frac{1}{2n+2}$ since if we start from $\frac{1}{2n+2}<\frac{1}{n+1}=\frac{2}{2n+2}<\frac{1}{2n+1}+\frac{1}{2n+2}→\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}>0$ Thus its positive. Oh yeah I saw the hint. Since its bounded and monotone its convergent by the monotone convergent theorem.

24. Jul 6, 2013

### Dick

Ok, I find that a little hard to follow, but it's probably ok. Yes, $y_n$ is increasing and bounded above. So it converges.

25. Jul 6, 2013

### bonfire09

Thanks so much. I think Im going to practice on other series and sequences until I get a good hang of it. I don't think its that bad now.