Question

A killer problem. If \(a_0=1\) and:\[a_n=a_{\frac{n}{2}}+a_{\frac{n}{3}}+a_{\frac{n}{6}}\]Find the convergence or divergence of \(\frac{a_n}{n}\).

Answer 1

This is complex analysis, by the way.

Answer 2

Or real. I actually have no idea.

Answer 3

I'm not sure I understand the notation....
\[a_{\frac{n}{2}}\]
only makes sense for an even "n"...

Answer 4

Yes, therefore it's only defined that \(n\mid\left(\frac{n}{2},\frac{n}{3}\right)\in\mathbb Z\).

Answer 5

this would diverge cuz its a harmonic series 1/n diverges i would use the comparison test, and conclude that the overall series diverges cuz ur comparing it with 1/n with diverges cuz p<1 i hope im right lol

Answer 6

It still doesn't make sense to me. The first term that would work is
\[a_{\frac{n}{6}} = a_3 + a_2 + a_1 \]
And we don't know any of those terms on the right.

Answer 7

My question as well; but the problem is correctly written, so!

Answer 8

Also, it's not harmonic. It's way too erratic.

Answer 9

lol well idk then, what class is this for? lmk when u find an answer im curious

Answer 10

Is this problem from a textbook or a problem set made up by a professor?

Answer 11

Setup by a professor. This problem was for chaos theory and stochastic analysis.

Answer 12

lol i havent even hit this yet, im still in calc 2 haha

Answer 13

Interesting. I apologize that I'm not of much help, I really just don't understand the way it's written or what exactly it's asking for. It doesn't make sense to me to have a sequence whose terms are only defined for certain values of n.

Answer 14

I don't know if @JamesJ is around right now but if so, maybe he can take a look. I might just be crazy or overlooking something.

Answer 15

It's a very weird problem.

Answer 16

Not well posed, but clearly divergent. If the new term is the sum of three (if they exist) previous positive terms, the new terms keep getting bigger. Hard to converge if it does this.

Answer 17

It's \[\frac{a_n}{n} \]

Answer 18

I think I'm going in the right direction for proof of existence of convergence of \(\frac{a_n}{n}\):\[f(x)=6x\mid\forall\left(0<x<\frac{1}{6}\right),f(x)=1\mid\forall\left(\frac{1}{6}\leq x<1\right)\text{; by define}\\f(x)=a_n\mid\forall\left(n\leq x<n+1\right)\text{; provisional}\\f(x)=f(\frac{x}{2})+f(\frac{x}{3})+f(\frac{x}{6})\mid\forall\left(x\geq1\right)\text{; another define}\\g(y)=f(e^y)e^{-y}\mid\forall\left(x\in\mathbb R\right)\text{; define; obviously monotonically bounded}\]If we can prove that \(\lim_{y\to\infty}g(y)\in\mathbb R\). Then, a Fourier transform at \(0\) and Tauberian theorem for proof. I'm still working on the last part. :D

Answer 19

Alternatively, it might be possible to prove that this system is not mappable in any space.