# Proving the accumulation point for a given sequence (using floors)

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 $$n$$.

• So I understand how you have used mods to show this, but how would I do this in terms of showing a vigorous analysis proof? Does this show that $s_n$ converges to those values? Sorry, I'm new to analysis.
– Bar
Commented Feb 2, 2020 at 23:10
• @Aas There's actually no convergence involved, as such, as the values are not changing (or, as I explain, it's actually a degenerate form of convergence). What I've shown, and you can also confirm by manual checking if you wish, is that the only values you get are $\frac{10}{11}$ for odd $n$ and $\frac{1}{11}$ for even $n$. This is basically a degenerate case of proving convergence, in this case separately for even & odd values, where the sequence values in those sub-sequences are not changing and, thus, their limit is their actual values. Commented Feb 2, 2020 at 23:14
• Oh, wow! That makes a lot of sense. I guess I got so caught up in proving using special theorems that I forgot the basics. Thanks so much!
– Bar
Commented Feb 2, 2020 at 23:19
• @Aas You're welcome. I suspect that at least one of the points of the exercise it to show the more complicated & powerful math techniques like limits, accumulation points, etc., also apply to quite simple cases, in addition to the more difficult cases you might normally deal with. Commented Feb 2, 2020 at 23:22