# 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
Feb 2 '20 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. Feb 2 '20 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
Feb 2 '20 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. Feb 2 '20 at 23:22