Does the fact that $\lim_{n\to\infty}[f(n)-f(n-1)]=0$ imply that $\lim_{n\to\infty}f(n)$ exists?

Question

I have a function $f(n)$ that is defined when $n$ is a non-negative integer. $f(n)$ is always a real number such that $1<f(n)<2$. Because of how this function is definied, I cannot prove the limit directly, but I was able to prove that $$\lim_{n\to\infty}[f(n)-f(n-1)]=0$$It makes sense to me that this might prove the existence of $\lim_{n\to\infty}f(n)$, but I do not know for sure if that is true. Is there some way to prove that this is true? And if so, is the fact that $f(n)$ is bounded between 1 and 2 necessary to the proof?

Answer 1

HINT: No. Let us think about a condition $0\leq f(n)\leq1$ (we can move and scale the values). Then we start:

up: $f(1)=0$, $f(2)=1$,

down: $f(3)=1/2$, $f(4)=0$,

up: $f(5)=1/4$, $f(6)=2/4$, $f(7)=3/4$, $f(8)=4/4=1$,

down: $f(9)=7/8$, $f(10)=6/8$, $f(11)=5/8$, $\ldots$. Can you see the pattern?

Answer 2

Nope. Example:

$\sin(2\pi x)$.