Harmonic numbers converging to any number

Question

$x_n$ 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?

Answer 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 $c_n=\pm 1$?