So the question is as given: Question
So basically, I need to show that the sequence has 2 accumulation points. For even terms, the accumulation point is 0.909 or 1/11 and for odd terms the sequence accumulates to 10/11. So, so far, I have created two sequences, one for even terms: s_{2n} and one for odd terms: s_{2n+1}. I seem to be having lots of trouble actually proving that the two subsequences converge given the fact that we have "floors" in the equation.
I'm thinking I could try and prove that both subsequences converge by showing that the limit definition holds OR by using induction.
Edit (If you cannot see the photo):
So this is the question:
$s_n = \frac{10^n}{11} - \lfloor \frac{10^n}{11} \rfloor$
Now prove that this has an accumulation point basically.
So what I've come up with is that for even sequences, $s_{2n}$, we have an accumulation point of $\frac{1}{11}$ and for odd sequences, $s_{2n+1}$ we get an accumulation point of $\frac{10}{11}$. The problem is that I need to prove this rigorously.
So, If I prove that the two subsequences converge to the numbers I've said above, then the overall sequence $s_n$ will have those fractions as accumulation points. So this is where I'm stuck. I've tried induction, and I've tried to use the epsilon-delta proof but I am really unsure what to do since I don't seem to understand how to get rid of the "floors" or just how to use them in the proof in general.
The sequence is
$$s_n = \frac{10^n}{11} - \left\lfloor \frac{10^n}{11} \right\rfloor \tag{1}\label{eq1A}$$
Note for all $x \in \mathbb{R}$ that $x - \lfloor x \rfloor$ is the fractional part of $x$ (e.g., if $x = n + r$ where $n \in \mathbb{Z}$ and $0 \le r \lt 1$, then $x - \lfloor x \rfloor = (n + r) - n = r$). Thus, the value of $s_n$ is the fractional part of $\frac{10^n}{11}$. In particular, it is the integer remainder of $10^n$ divided by $11$, when it is itself divided by $11$. You have
$$10 \equiv -1 \equiv 10 \pmod{11} \implies 10^{n} \equiv (-1)^n \pmod{11} \tag{2}\label{eq2A}$$
Thus, for odd $n$, you have $10^{n} \equiv -1 \equiv 10 \pmod{11}$, i.e., has a remainder of $10$, and for $n$ is even, you have $10^{n} \equiv 1 \pmod{11}$, i.e., has a remainder of $1$. As such, you get $s_n = \frac{10}{11}$ for all odd $n$, and $s_n = \frac{1}{11}$ for all even $n$.