Finding the cluster points of sequences So, I've been doing these questions and I've done 3 out of 4 (I think :/) but I'm not sure how to do the last one. Here's the question:
Find the cluster points and name one convergent sub-sequence of each of the following sequences. For this problem you don't need to prove your statements.
(a) $1, 1/2, 1, 1/3, 1, 1/4, 1, 1/5$,...
So I said;
${{a_n}{_k}}$ (That's a double sub-script) = 1,1,1,1,1,1,...
$\lim_{n\to\infty}$ $a_n{_k} = 1$
${a_n{_k}_{+1}}$ (Double sub-script) = $1/1, 1/2, 1/3, 1/4 = 1/n$
$\lim_{n\to\infty}$ $a_n{_k}_{+1}$ (Double sub-script) $= 0$
Therefore the two subsequences converge to 1 and 0 and consequently the cluster points of the sequence ${a_n}$ is 1 and 0.
(b) $(a_n)$, where $a_n = 1+\frac{1}{n^2}$ for all $n \in \mathbb{N}$ 
For this one I just did the limit of $a_n$ and got 1, so I concluded that the cluster point is 1
(c) $(a_n)$, where $a_1 = 5$, $a_{2k} = 2+\frac{1}{2k}$ and $a_{2k+1} = 6-\frac{1}{2k+1}$ for $k \in \mathbb{N}.$ Thus, the sequence is:
$5, 2\frac{1}{2}$, $5\frac{2}{3}$, $2\frac{1}{4}$, $5\frac{4}{5}$,...
For this one I said there were three subsequences, ($a_1, a_{2k}, a_{2k+1}$). Foudn the limits of each and got the cluster points were 6, 2 and 5.
(d) $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1...$
I'm stuck on this one so I'm unsure what to do, any help on this would be GREATLY appreciated also if you could check my previous solutions to see if I've done them correctly, or if there any errors in it. Thanks!! :)
 A: Your answers to $a$ and $b$ are correct, good job.
In the answer to $c$, $5$ is not a cluster point. Which subsequence converges to $5$? Remember : the cluster points of a sequence do not change if we remove finitely many terms from the start of the sequence, because the definition of a cluster point (or limit point) goes like : $\forall \epsilon > 0, \exists N \in \mathbb N, \forall n > N$ etc. , here the points under scrutiny are $a_n$, where $n>N$ will be some possibly arbitrarily large number if $\epsilon$ is chosen small. 

In short, the tail of the sequence matters as far as cluster points are concerned, not the head. So the answer is only that $2,6$ are the cluster points. Furthermore, a subsequence is itself a sequence, so $a_1$ is a term of the sequence $a_n$, not a subsequence. To give a subsequence, you have to specify infinitely many terms with distinct indices, which is not the case with $a_1$.

Similarly, you can approach $d$. This is a little more subtle problem, but is nevertheless very interesting.

Claim : Every positive integer is a cluster point of this sequence.

Proof : Every positive integer appears infinitely many times in the sequence, because it will appear first when we write down the list $1,2, ...,k$, then again when we write down $1,2, ..., k+1$, and similarly for any $n>k$, since the list of terms $1,2,...,n$ is contained in the subsequence, $k$ will be contained in the sequence infinitely many times. So just remove this subsequence, and you will see that $k$ is a cluster point of the sequence for every $k$ being an integer.
Now, this I leave you to see. Use the $\epsilon-\delta$ definition, try playing the contradiction game, but the end result is:

A sequence of integers can only have limit points which are integers.

In case you do not wish to solve this, here is a spoiler (hover over the yellow box):

 Suppose $x$ is not an integer and is a limit point of the sequence, let $\tau$ be the distance  of $x$ from the nearest integer, then it is greater than zero. If $x$ is a cluster point, then for $\epsilon = \frac \tau 4$, there should exist $N$ such that if $n>N$, then $|a_n - x| < \epsilon$. But then, $a_n$ are integers, and by the definition of $\tau$, this cannot happen!

These two points together show that the set of limit points is exactly the st of positive integers. 
A: For (d) every integers  appear infinitely many times.So,the trick is take constant subsequence for each integer.Like,
For 1,
$a_{n_{k}}=1,1,1,1,...$ which converges to 1.Similarly,you can find out that every integer is a cluster point of this sequence.    
