How to prove the divergence of the highest power of $2$ that divides $n$ -sequence

Question

My guess is that this sequence diverge to infinity since every now and then the power of $2$ number increases in the long run. I.e. for $2^x$, $x$ increases. Is it easy to show and prove this sequence diverge to infinity? I dont know where to start and how to do it step by step.

Highest power of $2$ that divides $n$: $$\{{1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,16,1,2,1,4,...\}}$$

Information about the sequence can be found on A006519.

Answer 1

Let's call the sequence $a_n$.

It's easy to show it diverges, for the subsequence $$ a_1, a_3, a_5, \ldots, a_{2k+1}, \ldots $$ is $$ 1, 1, 1, 1, \ldots, 1, \ldots, $$ which converges to $1$, while the subsequence $$ a_2, a_6, a_{10}, \ldots, a_{2 + 4k}, \ldots $$ is $$ 2, 2, 2, \ldots, 2, \ldots $$ which converges to $2$. But in general, if a sequence converges to a limit $L$, then every subsequence also converges to $L$. In other words, if your sequence converged to some limit $L$, then that limit would have to be both $1$ and $2$, which is impossible.

Since the definition of "diverges" is "doesn't converge," your sequence diverges.

Your observation that for every natural number $k$, we have $$ a_{2^k} = 2^k $$ does in fact mean that as $n$ goes to infinity, there is no upper bound on $a_n$. If you want to say that this means it "diverges to infinity," I suppose you're welcome to do so, but most mathematicians will not recognize that phrase as meaningful.

Answer 2

"Diverging" is the negation of "converging". In other words, if a sequence does not "settle on a finite value", then it diverges. As you have pointed out already, at any point in your sequence that corresponds to a pure power of $2$, the value of the sequence just gets higher and higher. So it clearly doesn't settle anywhere, and it is therefore divergent.

"Diverging to infinity" means it carries off to infinity, and also (just as importantly) that it doesn't come back down. However, in your sequence, there is a $1$ in every odd-numbered spot. So your sequence does not diverge to infinity.

The sequence is unbounded, though. That's a different name for a sequence that goes off to infinity (either positive or negative infinity), but this name still applies even if the sequence happens to come back down to small values from time to time.

Answer 3

This sequence diverges yes, but not to infinity. You can't fit the definition of it:

$$\lim_{n\to \infty}=\infty\iff\forall\delta>0, \exists n_0\in\mathbb{N}: n > n_0\Rightarrow u_n>\delta$$

However you can state that a sub-sequence of the sequence you defined diverges to infinity. In particular the sub-sequence $u_{2^k}, k\in \mathbb{N}$.