Solutions

(a) Notice that

$$\frac{2n + 1}{5n + 4} = \frac{2}{5} - \frac{3}{5 (5n + 4)}.$$

To obtain this result, simply divide $2n + 1$ by $5n + 4$, $\frac{2}{5}$ times. Thus,

$$\begin{aligned} \abs{\frac{2n + 1}{5n + 4} - \frac{2}{5}} = \abs{\frac{2}{5} - \frac{3}{5(5n + 4)} - \frac{2}{5}} = \frac{3}{5(5n + 4)} < \epsilon &\implies n > \frac{3 - 20\epsilon}{25\epsilon}. \end{aligned}$$

For any $\epsilon > 0$, by choosing $N = \frac{3 - 20\epsilon}{25\epsilon}$, if $n \geq N$, then

$$\abs{\frac{2n + 1}{5n + 4} - \frac{2}{5}} < \epsilon.$$

(b) Notice that

$$\forall n \in \mathbb{N}, \quad \abs{\frac{2n^2}{n^3 + 3}} < \abs{\frac{2n^2}{n^3}} = \frac{2}{n}.$$

Thus, if

$$b_{n} = \frac{2}{n}$$

converges, then so does

$$a_{n} = \frac{2n^2}{n^3 + 3}.$$

Using this trick, the rest of the proof is left to the reader.

(c) Similarly to (b), we notice that our sequence is bounded above:

$$\forall n \in \mathbb{N}, \quad \abs{\frac{\sin(n^2)}{\sqrt[3]{n}}} \leq \abs{\frac{1}{\sqrt[3]{n}}}.$$

Again, by proving that

$$b_{n} = \frac{1}{\sqrt[3]{n}}$$

is convergent, we will have succeeded in proving that

$$a_{n} = \frac{\sin(n^2)}{\sqrt[3]{n}}$$

is convergent as well (why?). With this hint, the rest of the exercise is left to the reader.