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!! :)

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.    
