### 9.11.05

## annoying animated gifs

last night at about 3am, i discovered that the sum to infinity of the nth Fibonacci number divided by two the nth power is .. 4!

I was quite interested and disappointed at the same time. Interested because I didn't know it (although I should have). Disappointed because if it had been less than four, it would have helped me out quite a bit.

The result is not too groundbreaking; if you know what a generating function of a sequence of integers is, you can work it out, plug in 1/2 and get 4.

You can also find it out some simpler ways; the way I found it is close to an obvious calculation you might make if you know a little about infinite sums.

I was trying to construct a variation on the Cantor set where instead of removing the middle third of each interval, you remove the third quarter:

Looking at it that way, it's reasonably clear that the black set will end up having the same length as the original interval.

If you like, the area of the white set will be

3/4 * 3/4 * 3/4 * 3/4 * ... = 0, since each subsequent white set is three-quarters the length of the previous one.

But if you group the removed (black) intervals by their size, there are F

So their total length is 1/4 times the sum to infinity of 1/2

I'd been working this out in the hope that the white set would have a positive length, which would have been useful to me. But that's another story.

I was quite interested and disappointed at the same time. Interested because I didn't know it (although I should have). Disappointed because if it had been less than four, it would have helped me out quite a bit.

The result is not too groundbreaking; if you know what a generating function of a sequence of integers is, you can work it out, plug in 1/2 and get 4.

You can also find it out some simpler ways; the way I found it is close to an obvious calculation you might make if you know a little about infinite sums.

I was trying to construct a variation on the Cantor set where instead of removing the middle third of each interval, you remove the third quarter:

Looking at it that way, it's reasonably clear that the black set will end up having the same length as the original interval.

If you like, the area of the white set will be

3/4 * 3/4 * 3/4 * 3/4 * ... = 0, since each subsequent white set is three-quarters the length of the previous one.

But if you group the removed (black) intervals by their size, there are F

_{n}intervals of length 1/2^{n+2}, on an interval of length 1. (F_{n}the nth Fibonacci number, of course)So their total length is 1/4 times the sum to infinity of 1/2

^{n}* F_{n}, which is the same as the length of the interval, ie 1.I'd been working this out in the hope that the white set would have a positive length, which would have been useful to me. But that's another story.