Proving divergence for a tricky series

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Homework Statement



This problem takes a bit of background, so please bear with me.

The assignment reads: Suppose you have a large supply of books, all the same size, and you stack them at the edge of a table, with each book extending farther beyond the edge of the table than the one beneath it. Show that it is possible to do this so that the top book extends entirely beyond the table. In fact, show that the top book can extend any distance at all beyond the edge of the table if the stack is high enough. Use the following method of stacking: The top book extends half its length beyond the second book. The second book extends a quarter of its length beyond the third. The third extends one-sixth of its length beyond the fourth, and so on.

Here is a graphic of what is being described:

tQNp2.jpg


Here is the series that the professor shows us to be what models this sort of thing:
6ykIo.jpg


Homework Equations



Well, I feel the divergence test, the comparison test, and the integral test are all relevant.


The Attempt at a Solution



This is where it gets a bit messy. I am still not entirely sure what is being asked of me. I know that divergence comes into play, and I know that the series diverges. The way that this problem is presented is: The professor goes through and shows what sort of series models this, tells us that it diverges, shows us that the books will never fall, and asks us to explain why.

Unfortunately, I'm not entirely sure what to do. Prove divergence? The professor gave us this as a hint: "This is related to the harmonic series as well as 1/2n. You must use a test to show it diverges. The test you use will establish that the partial sum grows without bounds. You need to conclude that for any finite distance from the table, there is a finite value n, where the partial sum gets the stack past the distance."

So I understand that somewhere in my report I need to prove divergence. But I am just not sure how, and I am a bit fuzzy on if there is other information being asked of me. This is required to be in a report style format, but I am finished with everything else besides this part.

Help much appreciated and thank you for reading :)
 
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Is this not just the harmonic series?
 
From the images I gather that your series in this one:

<br /> c=const.<br />

<br /> \sum_{n=1}^{\infty}\frac{c}{2n}<br />

now you can use any convergence/divergence test you know :)
 
gomunkul51 said:
From the images I gather that your series in this one:

<br /> c=const.<br />

<br /> \sum_{n=1}^{\infty}\frac{c}{2n}<br />

now you can use any convergence/divergence test you know :)

How would I establish that the series that I have can be compared to that one? I can see the similarity, but I feel like if I just introduced that series out of no where that I would lose points. Do I manipulate it? Or is that series smaller than the one that I have, where I could define it as BsubN and do a comparison test between it and my original series?
 
Mohdoo said:
How would I establish that the series that I have can be compared to that one? I can see the similarity, but I feel like if I just introduced that series out of no where that I would lose points. Do I manipulate it? Or is that series smaller than the one that I have, where I could define it as BsubN and do a comparison test between it and my original series?
It is written clearly in the second image as L*(0.5 + sum).
the sum is the only part that can diverge, others are constants.

and the sum is basically (with some constant multipliers) the harmonic series (1/n) as stated by hunt_mat.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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