0

$x_n$ is equal to $x_{n-1}-\frac 1n$ if $x_{n-1}+\frac 1n \ge a$ and is else equal to $x_{n-1}+\frac 1n$ where $a$ is any real number and $x_0$ is a rational number (its usually defined by $\lfloor a \rfloor$). Does that mean that $$\lim_{n\to \infty} x_n = a?$$ And how fast does it converge to $a$? You could actually write $x_n$ as $\displaystyle\sum_{k=1}^n \dfrac {c_k}k$ where $c_k$ is either $-1$ or $1$.

I was also wondering if its possible to replace $\dfrac 1n$ with any other function that converges to zero. If so, does that mean that the summation of this function (like the harmonic series $\frac 1n$) needs to diverge?

JohnColtraneisJC
  • 1,919
Kinheadpump
  • 1,331

1 Answers1

1

Some broad hints, to let you keep exploring on your own:

  1. For simplicity's sake, I'll take $a\gt 0$. Because you know the harmonic series diverges, there's some $N$ such that $\sum_{n=1}^N\frac1n\gt a$; let $N_0$ be the first such $N$. Then $x_{N_0-1}\lt a$ and $x_{N_0}\gt a$; what does that imply about how far $x_n$ can be from $a$? You can repeat this argument every time your $x_n$ changes 'sides'...
  2. How far apart are $x_n$ and $x_{n+1}$? Can you see what that implies about how far one of them must be from $a$? That should answer your question about speed of convergence.
  3. On the subject of replacing $\{\frac1n\}$ with some other sequence $\{s_n\}$: Suppose that $\sum_n \left|s_n\right|$ converges to some number $\alpha$. Can you express any number $\beta\gt\alpha$ as the sum $\sum_nc_ns_n$ where $c_n=\pm 1$?
Steven Stadnicki
  • 51,746
  • (To 1) But with my method, $x_n$ never changes sides – Kinheadpump Aug 20 '19 at 18:36
  • Ahh - I misread, apologies! I thought you were 'overshooting' and then backstepping; that is, always adding/subtrracting the next term to go in the direction of a. The arguments still go through largely unchanged, though; since $\sum_n\frac1n$ diverges, you can show that there's always a 'backstep', you can show a bound on how far $x_{N_0}$ can be from $a$ at such a step, and you can show that there's always a next backstep. – Steven Stadnicki Aug 20 '19 at 19:05
  • @RossMillikan If you read more carefully, the definition in the $a=\sqrt{3}$ case would be that $x_1=1$, $x_2=\frac32$, and then since $\frac32+\frac13 = \frac{11}6\gt\sqrt{3}$, instead we take $x_3=\frac32-\frac13=\frac76$. Now $\frac76+\frac14=\frac{17}{12}\lt\sqrt{3}$, so $x_4=\frac{17}{12}$, etc. But the same core argument still goes through, since if $x_n\lt x_{n-1}$ then it's because $x_{n-1}+\frac1n\gt a$, so we can bound $\left|x_n-a\right|$ by $\frac2n$, etc. – Steven Stadnicki Aug 20 '19 at 22:25