prove using the definition of the convergence of a sequence that the limit as n approaches infinity of sin(n^2)/(n^(1/3))=0 and so it converges.
So I know that a sequence (an) to converges to a real number a if, for every positive number Є, there exists an N ∈ N such whenever n≥N it follows that |an-a|< Є.
I don't understand how to exactly find an epsilon that works for this sequence.
Best Answer
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 $\epsilon$.