Proving this series converges or diverges

[bonfire09]

Homework Statement

Prove y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} where n\in\mathbb{N} is convergent or divergent.

Homework Equations

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.

 

[Mandelbroth]

Homework Statement

Prove y_n=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} where n\in\mathbb{N} is convergent or divergent.

Homework Equations

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.

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

 

[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.

 

[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)

 

[Mandelbroth]

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.

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?

 

[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}

 

[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

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

 

[bonfire09]

so is the way I have it not correct then?

 

[lurflurf]

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

 

[Dick]

so is the way I have it not correct then?

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

 

[bonfire09]

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

 

[Dick]

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

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?

 

[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.

 

[Dick]

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.

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.

 

[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

 

[Dick]

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.

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.

 

[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.

 

[Dick]

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 its convergent.

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.

 

[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.

 

[Dick]

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.

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?

 

[bonfire09]

Well \frac{-1}{n+1}&lt;\frac{1}{2n+1}+\frac{1}{2n+2} since if we start from \frac{-3}{n+1}&lt;\frac{2}{2n+1}=\frac{-2}{n+1}&lt;\frac{2}{2n+1}+\frac{1}{n+1}=\frac{-2}{n+1}&lt;\frac{2}{2n+1}+\frac{2}{2n+2}=\frac{-1}{n+1}&lt;\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.

 

[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.

 

[bonfire09]

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

 

[Dick]

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

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.

 

[bonfire09]

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

 

[Millennial]

Just a note, the value the sequence converges to can be found this way:

\lim_{n\to\infty} \sum_{k=0}^{2n} \frac{1}{k} - \sum_{k=0}^{n} \frac{1}{k}<br /> = \lim_{n\to\infty} \sum_{k=0}^{2n} \frac{1}{k} - \log(2n) - \sum_{k=0}^{n} \frac{1}{k} + \log(n) + \log(2)<br /> = \gamma - \gamma + \log(2) = \log(2)

 

[Dick]

Just a note, the value the sequence converges to can be found this way:

\lim_{n\to\infty} \sum_{k=0}^{2n} \frac{1}{k} - \sum_{k=0}^{n} \frac{1}{k}<br /> = \lim_{n\to\infty} \sum_{k=0}^{2n} \frac{1}{k} - \log(2n) - \sum_{k=0}^{n} \frac{1}{k} + \log(n) + \log(2)<br /> = \gamma - \gamma + \log(2) = \log(2)

Yeah, that's what lurflurf was pointing out. I think it's more elementary to approach it as a Riemann sum as in post 10.