1

On our last Complex analysis course, our professor announced his retirement. Upon ending the class, he mentioned that he has an interesting problem he wants to leave me with, given my interest in analysis. He said that he was trying to solve this for the last $30$ years but didn't manage to come to a conclusion . Here is the problem :

Let $x_n$ be a sequence such that: $$x_n = n^2sin(n)$$

What's the limit of $x_n$ as $n \rightarrow \infty$ ?

I assure you, this is the full text of the problem and nothing has been omitted . I'm sorry if I show signs of poor understanding, but does posing such a question even make sense ? Is there even a point in starting to try on solving this problem ?

The alternating signs of the $sin$ function and unpredictable behaviour of the sequence makes me wonder if there's any answer to be found at all.

Victor
  • 3,213
  • Exact limit seems like odd wording. Maybe set of limit points is intended. – André Nicolas Jun 23 '15 at 16:10
  • Sorry, we use this kind of wording quite often in my language, it didn't occur to me it might cause confusion. The problem simply requires the limit of the sequence as $n \rightarrow \infty$ – Victor Jun 23 '15 at 16:16
  • 1
    I assumed that one could not possibly mean limit, since the fact that the limit does not exist takes a few minutes to see, not $30$ years. – André Nicolas Jun 23 '15 at 16:45

1 Answers1

2

I don't know if this is what you're looking for, but one thing we can say for certain is that the sequence

$$ x_n=n^2\sin(n) $$

does not converge to a limit as $n$ converges to infinity. Indeed, suppose it did. Then we could multiply by the convergent sequence $1/n^2$, and conclude that the sequence $\sin(n)$ converges, which isn't the case.

Alternatively, by Dirichlet's approximation theorem, we can find sequences $p_n,q_n$ of integers such that

$$ q_n\pi-p_n\to 0 $$

Then we get

$$ \sin(q_n\pi-p_n)=\sin(q_n\pi)\cos(p_n)-\cos(q_n\pi)\sin(p_n)=(-1)^{q_n+1}\sin(p_n) $$

We conclude that $\sin(p_n)\to0$. Now consider:

$$ \sin(p_n+1)=\sin(p_n)\cos(1)+\cos(p_n)\sin(1)\to\sin(1) $$

So the sequence $\sin(n)$ has a subsequence that converges to a positive value, and therefore the sequence $n^2\sin(n)$ diverges.

Community
  • 1
John Gowers
  • 24,959