A Fun Series Problem

Added: 2026-05-25

[00:00:00]Okay, so we're going to evaluate this series. I think just looking at it, it's not particularly obvious what sort of approach we could take. So it's one where seeing this fresh for the first time, you might just get stuck in and explore doing some algebraic manipulations if you can find anything that might help simplify the problem. I think if we look at this function that we're summing, 2 to the n over This is really interesting, this 2 to the 2 to the n term in there. And we could actually think about this in terms of what happens if we were to square this term. So we had 2 to the 2 to the n squared. So if we multiply this by itself, then our power is being multiplied by 2. So we're multiplying this power, 2 to the n * 2. So then the new power becomes 2 to the n + 1 because 2 to the n * 2 is the same as 2 to the n + 1. So it's perhaps a bit strange working with indices where we've got three layers of indices there rather than just two. And this is particularly interesting now because if we're thinking about squaring this, we could actually try rewriting the denominator as a difference of two squares kind of expression. So multiplying on the top and the bottom of the fraction by 2 to the 2 to the n - 1, this doesn't change the value of the fraction. But then this gives us, if we leave the numerator alone for now, as we've still got 2 to the n * 2 to the 2 to the n - 1. But then this difference of two squares expression, as we know what the square of this 2 to the 2 to the n term is, we get 2 to the 2 to the n + 1 - 1. I think this is really interesting now. We've got 2 to the 2 to the n + 1.

[00:01:32]And here in the denominator we've got 2 to the 2 to the n. So this is suggesting that perhaps we could actually even try and split this fraction up further and go for a method of differences telescoping series kind of approach. So we could aim for some sort of denominator over 2 to the 2 to the n - 1. And if we could set this up so that then we subtract some sort of other term divided by 2 to the 2 to the n plus 1, the next term or minus 1, then we could set this up so that we add this term, we add the next term. There could be some really nice cancellation between the n plus 1 term and then that would become the n plus 1 term and the next one. So, let's see what this looks like and see if we can make this work. So, what we really be aiming for is somehow rewriting this fraction. You can see we've got the denominator 2 to the 2 to the n plus 1. We want to split this up so there's also a piece that has a denominator just 2 to the 2 to the n where we don't have the n plus 1. And we could actually achieve this by thinking of this as 2 to the 2 to the n plus 1 if we were able to write the numerator somehow as having 2 to the 2 to the n plus 1 and then divide this by 2 to the 2 to the n plus 1 minus 1. Think about factorizing the denominator now. We've got 2 to the 2 to the n plus 1 in the numerator divided by and then rewriting this difference of two squares, just refactorizing it like we had earlier, 2 to the 2 to the n plus 1. Those two terms are going to cancel and we'll just be left with 2 to the 2 to the n minus 1 and then our numerator would just become 1. So, these two brackets would entirely cancel. So, we could try and aim for somehow getting this 2 to the 2 to the n plus 1 in our numerator somehow. And we can actually do this because if I just call this star and then we've got a lot of rough work in the middle there. So, if we say this star term now, this is going to be equal to we could get a 2 to the 2 to the n plus 1 term really easily just by adding 1 and then we'd have to subtract 2 to compensate to make a negative 1 there and we'll leave this 2 to the n term on the outside. So, we've got 2 to the n times 2 to the 2 to the n plus 1 minus 2 all like this. You see we haven't changed anything there. We've just got plus 1 minus 2 instead of a minus one.

[00:03:53]And then this is still being divided by two to the two to the n plus one minus one.

[00:03:59]And we can rewrite this as two separate fractions. So, we've got two to the n two to the two to the n plus one. And then we'll take this negative two into a separate fraction. It's still being multiplied by this two to the n term.

[00:04:10]So, take away two to the n times two.

[00:04:13]So, this is just two to the n plus one effectively.

[00:04:17]And our denominator here is still two to the two to the n plus one minus one. It's the same here as well. Two to the two to the n plus one minus one. We've just seen that this term actually cancels really nicely. You can see here we've got two to the two to the n plus one over two to the two to the n plus one minus one, which is exactly the same as this fraction we've got here. So, we know that this actually cancels to one over two to the two to the n minus one. So, then we're just left with one times two to the n. So, we've got two to the n over and then all that effectively happens is this n plus one is reduced into just an n. So, now we've got two to the two to the n minus one. And then our second one becomes two to the n plus one over two to the two to the n plus one minus one. Then you can see we're setting ourselves up really nicely for a really elegant telescoping kind of argument now.

[00:05:17]And just so that we're being rigorous and careful with this, we'll rewrite this sum to infinity as a limit. So, this is really the limit as let's say capital N goes to infinity of our sum from zero up to this new capital N of the same terms two to the n over two to the two to the n minus one minus two to the n plus one over two to the two to the n plus one minus one. And if we just write out the first couple of terms, it'll become clear exactly how this cancellation structure is going to work from one term to the next. So, we've got the limit as capital N goes to infinity of our first term when lowercase n is zero, we've got 2 to the zero, which is just 1 over 2 to the 2 to the zero. So, this is just 2 to the power of 1 minus 1. Then our next term, we've got 0 plus 1, so this becomes 2 to the power of 1, or just 2, over 2 to the power of and again it's 2 to the n plus 1 gives us a 2 and take away 1. So, so far nothing's canceling, but then if we go on to the next term where n is equal to 1, we've got plus 2 to the 1, so this gives us a 2 over 2 to the 2 to the 1, so 2 squared minus 1.

[00:06:31]You can see these two terms cancel with each other. So, we've effectively just got our previous n plus 1 becomes our current n term effectively. And then the next one, we take away when n is 1, we've got 2 squared over 2 to the 2 squared minus 1. And you can see this is going to cancel with our next term cuz then when n is equal to 2, we've got 2 squared over 2 to the 2 squared minus 1. And then the next term and so on, they're all going to cancel in pairs. So, this term cancels with this, this term cancels with this and so on.

[00:07:07]So, we've got our first term, but then we also need to think carefully about what happens at the end here, which is why we're taking limits. We do need to make sure that this sum is actually going to converge setting it up like this. So, our final term, we're going to have 2 to the capital N taking lowercase n equals capital N here over 2 to the 2 to the capital N minus 1. So, that term's going to cancel with the previous one in the sum from our n minus 1 term.

[00:07:33]So, that's going to cancel nicely. And then all we're left with is our 2 to the capital N plus 1 over 2 to the 2 to the capital N plus one minus one. So, what we're really left with are two terms now. So, we need to consider the limit as capital N goes to infinity of this term. One over two minus one is just the number one. So, we've got one minus two to the N plus one over two to the two to the N plus one minus one like this. So, I think this term, it's not super obvious how this limit's going to work or exactly how we could show that this term's going to go to zero as N goes to infinity. But, we could think of this as introducing, let's say, two to the N plus one equals X, for example. Then, we could rewrite all of this as we've just got X divided by two to the X minus one, which I think makes it much, much clearer that we've now got the numerator is just linear. It just grows like X as X goes to infinity, whereas the denominator is exponential, two to the power of X minus one. So, the denominator is going to grow to infinity so much faster than X that we can be confident this is going to zero as This is as X goes to infinity. You can see that as N goes to infinity, two to the N plus one certainly goes to infinity. So, we are interested in X going to infinity. So, actually, this term then as N goes to infinity, this just goes to zero. And then, we can conclude then that our original sum is actually just equal to one, which I think is really interesting and really elegant that we can use this structure and set up this really beautiful telescoping argument for something that looked really complicated and get such a nice, simple answer in the end using some of the right algebraic trickery.