Our sequence is $\dfrac{\sin (n^2)}{\sqrt[3]{n}}$.
Now, what we are going to do first, is the following:
note that $\dfrac{-1}{\sqrt[3]{n}} \leq \dfrac{\sin (n^2)}{\sqrt[3]{n}} \leq \dfrac{1}{\sqrt[3]{n}}$. Rewritten succinctly, $\Bigg|\dfrac{\sin (n^2)}{\sqrt[3]{n}}\Bigg| \leq \Bigg|\dfrac{1}{\sqrt[3]{n}}\Bigg|$
Let $\epsilon > 0$. Now, there exists $N \in \mathbb{N}$ such that $n > N \implies \Bigg|\dfrac{1}{\sqrt[3]{n}}\Bigg| < \epsilon$. Namely, we can let $N = \dfrac{1}{\epsilon^3}$, for example.
Therefore,
$$
n > N \implies \Bigg|\dfrac{\sin (n^2)}{\sqrt[3]{n}}-0\Bigg| \leq \Bigg|\dfrac{1}{\sqrt[3]{n}}\Bigg| < \epsilon
$$
Hence, we are done, with $a=0$.
What was important in this proof, was the use of the boundedness of the $\sin$ function. It allowed us to considerably simplify the calculation of $N$ given $\epsilon$.
Your main question seems to be
Why not define sequences on some other index set?
Okay, the term sequence is reserved for functions $a : \mathbb N \to X$ (note that I wrote $X$ instead of $\mathbb R$ because we may consider sequences in arbitrary sets, or if we are interested in convergence, in arbitrary topological spaces). However, this can easily be generalized to other index sets. This results in the concept of a net. See here. Quote:
Let $A$ be a directed set with preorder relation $\ge$ and $X$ be a topological space. A function $f: A \to X$ is said to be a net.
Recall that a preorder relation $\ge$ on a nonempty set $A$ is a reflexive and transitive binary relation. We do not require it to be antisymmetric (if it is, it is called a partial order). $(A,\ge)$ is called directed if for all $a, b \in A$ there exists $c \in A$ with $c \ge a$ and $c \ge b$. Simple examples are $\mathbb N, \mathbb Z, \mathbb Q, \mathbb R$ with their natural partial order. These sets are even totally ordered which means that for any two elements $a, b$ we have $a \ge b$ or $b \ge a$.
We often write nets in the form $(x_\alpha) = (x_\alpha)_{\alpha \in A}$. A net $(x_\alpha)$ is said to convergence to $x \in X$ if each open neigborhood $U$ of $x$ in $X$ admits $\alpha_0 \in A$ such that $x_\alpha \in U$ for all $\alpha \ge \alpha_0$. This obviously extends the concept of convergence from sequences to more general objects.
Interesting examples of nets which are no sequences occur in calculus when the Riemann integral is introduced. Here we consider partitions $\mathfrak{P} = (t_0, \ldots,t_n)$ of an interval $[a,b]$ and order them by"set inclusion" (i.e. $\mathfrak{P}' \ge \mathfrak{P}$ if $\{ t'_0, \ldots,t'_{n'} \} \supset \{ t_0, \ldots,t_n \}$). The uncountable set $P$ of all partitions is a directed partially ordered set, but it is not totally ordered. For any bounded function $f : [a,b] \to \mathbb R$ we consider the lower and upper Riemann sums which gives us two nets indexed by $P$. Both nets converge, but in general do not have the same limit. If they have, then $f$ is called Riemann integrable and the common limit is called the Riemann integral of $f$.
However, if you study calculus textbooks, you will find that frequently one again works with suitable sequences of partitions, e.g. equidistant partitions with mesh $(b-a)/n$. But note that conceptually one has to go beyond sequences.
An equivalent approach to the Riemann integral is by taking the elements of a modified index set $\mathbf P$ to be all systems $(\mathfrak{P},\mathbf x)$ where $\mathfrak{P} = (t_0, \ldots,t_n)$ is a partition of $[a,b]$ and $\mathbf x = (x_1,\ldots,x_n)$ with $x_i \in [t_{i-1},t_i]$. Defining $(\mathfrak{P}',\mathbf x') \ge (\mathfrak{P},\mathbf x)$ iff $\mathfrak{P}' \ge \mathfrak{P}$ we get a preorder on $\mathbf P$ which is directed but not antisymmetric. For any function $f : [a,b] \to \mathbb R$ we consider the Riemann sums $R(f;\mathfrak{P},\mathbf x) = \sum_{i=1}^n f(x_i)(t_i - t_{i-1})$. This system is a net over $\mathbf P$, and it is well-known that this net converges iff $f$ is Riemann integrable (and in that case the limit of the net is the Riemann integral of $f$). This shows once again that nets which look at first glance "fairly exotic" occur quite naturally in calculus and that we do need a concept of convergence for such nets.
So what is the reason for the prominent role of sequences in calculus? First of all, they are much simpler than general nets. For example, you can use induction. But the deeper reason is that all topological properties of $\mathbb R$ can be expressed via sequences. For example, $M \subset \mathbb R$ is closed iff for all sequences $(x_n)$ in $M$ which converge to some $x \in \mathbb R$ one has $x \in M$. Moreover, $M \subset \mathbb R$ is compact iff each sequence $(x_n)$ in $M$ has a convergent subsequence. Also the continuity of functions can be expressed via sequences: A function $f : \mathbb R \to \mathbb R$ is continous at $x$ iff for all sequences $(x_n)$ such that $x_n \to x$ one has $f(x_n) \to f(x)$.
Let me finally remark that series $\sum_{n=1}^\infty a_n$ should (at least in my opinion) be treated as nets. The usual approach is to consider the sequence $(s_m)$ of partial sums $s_m = \sum_{n=1}^m a_n$ and study its convergence. I suggest to introduce the set $\mathbf F$ of all finite subsets of $\mathbb N$ and order it by inclusion ($F' \ge F$ if $F' \supset F$). This produces a directed partially ordered set $\mathbf F$. Define a net over $\mathbf F$ by $(\sum_{n\in F}a_n)_{F \in \mathbf F}$. This net may converge or not. If it converges, we denote its limit by $\sum_{n\in \mathbb N} a_n$. It is now a nice exercise to show that $\sum_{n\in \mathbb N} a_n$ exists iff $\sum_{n=1}^\infty a_n$ is unconditionally convergent. Perhaps this unusual approach seems to be too exotic, but it has the benefit that it can be generalized to sums over uncountable index sets. Such sums occur in the context of Hilbert spaces. It is well-known that each Hilbert space $H$ has a (possibly uncoutable) orthonormal basis $\{b_\alpha\}_{\alpha \in A}$ and that each $x \in H$ can be written as $x = \sum_{\alpha \in A}\langle x, b_\alpha \rangle b_\alpha$.
Best Answer
I believe your question to be a language one, hence there is no difference in the formulation.