PROBLEMS

For each of the following sequences, determine if the sequence converges or diverges. If the sequence converges, determine the limit
a. $a_n=\frac{2^{n+1}}{3^{n+2}}$
b. $\quad a_n=\frac{(n+1)^2}{5 n^2+2 n+1}$
c. $a_n=\frac{\sin (n)}{n+1}$

a.
$$a_n=\frac{2^{n+1}}{3^{n+2}}=\frac{2}{9}\left(\frac{2}{3}\right)^n \rightarrow 0 \text { since } \frac{2}{3}<1$$

b.
$$a_n=\frac{(n+1)^2}{5 n^2+2 n+1}=\frac{n^2+2 n+1}{5 n^2+2 n+1}=\frac{1+\frac{2}{n}+\frac{1}{n^2}}{5+\frac{2}{n}+\frac{1}{n^2}} \rightarrow \frac{1+0+0}{5+0+0}=\frac{1}{5}$$

c.
$$a_n=\frac{\sin n}{n+1} \quad \text { Note }|\sin n| \leq 1$$
Thus $\left|a_n\right| \leq \frac{1}{n+1} \rightarrow 0$.
Thus $a_n \rightarrow 0$.

d. $a_n=\cos (n)$
e. $a_n=\frac{2(n+1)^2+e^{-n}}{3 n^2+5 n+10}$
f. $a_n=\frac{n \cos \left(\frac{n \pi}{2}\right)}{n+1}$

d. $a_n=\cos n$ does not approach a limit Thus the sequence diverges.

e.
$$a_n=\frac{2(n+1)^2+e^{-n}}{3 n^2+5 n+10} \rightarrow \frac{2 n^2}{3 n^2}=\frac{2}{3} \quad \text { since } e^{-n} \rightarrow 0$$

$\mathrm{f}$.
$$a_n=\frac{n}{n+1} \cos \left(\frac{n \pi}{2}\right) \rightarrow \cos \left(\frac{n \pi}{2}\right)$$
Since $\cos \left(\frac{n \pi}{2}\right)=0, \pm 1$, depending on $n$ the sequence diverges.

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