A Sequence Of Interest

This is just an interesting problem that came up the other day, and I rather enjoyed my solution to it.

We have a sequence,

Our goal is the value of as n goes to infinity.

For those interested, the first few values are approximately 1, 1.414, 1.848, 2.294, 2.749, ... The exact form of the values quickly becomes a mess of nested roots I wouldn't want to subject anyone to.

The first thing to do is a rather nice substitution, putting in the definition of to achieve



Several things become obvious right away. Most importantly, perhaps, is that the sequence is increasing. This is pretty clear from the numeric values I gave for the sequence, but this verifies it since



Giving



That in hand, I would like to generalize a bit and look at the function



Specifically, I want to think about its behavior as x goes to infinity. Just straight up plugging in infinity is rather unhelpful, since you get infinity - infinity, but with some cunning manipulation ... observe.











When I first wrote up this problem to turn it (it's for a class), once I got to the fraction representation, I was able to use L'Hopital's rule to solve this. As such, I titled my proof for Captain Bohman 'A Sequence of Interest : or How I Learned to Stop Worrying and Love L'Hopital'. Unfortunately, I then came up with the much nicer substitution you see above, which removes the need for L'Hopital, and thus ruins what is otherwise a great title. Just a personal side note.

But now we have a much nicer form of the expression, and we can do many things with it. For example,



Thus,



Which I find is a fascinating limit. Really it's why I like this problem so much.

But before we go back to the sequence, a couple of things to note. As x increases, is strictly decreasing, which means that is strictly increasing. Since at x = 0 the function has the value 0, it is strictly increasing, and the limit as x goes to infinity is 1/2, the function increases from 0, continuously passes through all values from 0 to 1/2, finally reaching 1/2 very far from now. Just keep that in mind.

And why is all this important? Because if we jump back to our sequence, and look at successive differences of terms, we get



Since, as we discussed, is increasing, the limit can be applied to say



And we can use facts about the limit to say nice things about those differences. For example, the successive differences are strictly increasing, and increase to 1/2. This is important, because it means that every difference is bounded from above by 1/2.

And that is what we can use do the following. Because 1/2 is greater than every successive difference, not that we can do the following












Since , we have a rather nice upper bound on ,



This will come in handy. Remember that one.

Now we want to bound from below. This gets a little trickier. Remember how increased smoothly from 0 to 1/2? As I pointed out, the successive differences do the same thing, though much less smoothly. That is to say, given some value between 0 and 1/2, we'll call it f, I can find an N such that for any n greater than N,



That is, any successive differences after N will be bound from below by f. We can then bound the sequence from below, like so. For convenience, I'm going to say that f equals p*(1/2), for some value of p between 0 and 1. Though obviously p is less than 1. So, given a value of p, I can find an N such that



Or, more nicely,



But remember, all successive differences there on are also bound by p*(1/2), so





Extrapolating outward, we get, for any positive integer k



And that gives us a lower bound, starting from N and going onwards.

Now we can combine our bounds to say, for a given value of p and the appropriately chosen value of N,



Now things start falling into place. We just just follow our noses. Start by dividing through by N + k.





With that in hand, we can start discussing what happens as k increases. Note that N is constant, so as k increases, is a constant divided by a larger and larger value, and it goes to 0. Related, as k goes to infinity, is essentially k/k, and will go to 1. On the other side, is a constant divided by an increasing value, so it will go to 0 as well. So as k increases to infinity, you're left with



Which is the same as saying, for exceptionally large n,



Which is quite nice! Basically we've squeezed the ratio between the ratios of the upper and lower bounds of the sequence, and squeezed the limit of the ratio to something between p*(1/2) and 1/2. Note though that p was said to be less than 1, so p*(1/2) is less than 1/2. So the limit in the bound above could actually be less than 1/2. However, notice that this derivation is valid for any value of p less than 1, and so the inequality above must be valid for any value of p less than 1. So I can essentially take the above bound, and squeeze p towards 1 as close as I want, which pushes the ratio closer and closer to 1/2 before finally nailing it there.

And we get



And so it's done.

Probably much more loquacious than need be, but I'd hate to think I lost anyone. Hope you enjoyed that as much as I.

*wag*