Proof for sequence $a_n$ with accumulation points $0$ and $2$

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I tried answering the following question, do my arguments make sense and are they correct?

Suppose we have a sequence $a_n$ with positive terms and accumulation points $0$ and $2$.

We consider the new sequence, $n\in \mathbb{N}_+$:

$$ b_n= \frac{n+a_n}{n \cdot a_n}$$

I am now asked to answer the questions:

(a) give an accumulation point of $b_n$

Well, notice that: $$ b_n= \frac{n+a_n}{n \cdot a_n}=\frac{1}{a_n} + \frac{1}{n}$$ In the limit for large $n$, we have that $\frac{1}{n} \rightarrow 0$. We notice that $a_n$ has a subsequence that converges to $2$, then certainly we have that $b_n$ has a subsequence $b_{n_j}$: $$ b_{n_j}=\frac{1}{a_{n_j}} + \frac{1}{{n_j}}\rightarrow \frac{1}{2}+0$$ So $\frac{1}{2}$ is an accumulation point.

(b) Is $b_n$ bounded?

Notice that we have subsequence $a_{n_k}$ that converges to $0$, we now consider $b_{n_k}$, such that: $$ b_{n_k}=\frac{1}{a_{n_k}} + \frac{1}{n_k}\rightarrow \infty $$ Because the sequence $b_n$ has a subsequence that is divergent to $\infty$, it cannot be bounded $\square$.

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The argumentation seems to be correct.

I would just caution you to be a little more careful with notation.

How is $a'_n$ being labeled? Do the subindices coincide when passing to $b'_n$?