Bounded sequences

a) The sequence is not constant since $a_1=-8$ and $a_2=-\dfrac{16}{3}$.

To verify if the solution is increasing or decreasing it is sufficient to verify whether $a_n\leq a_{n+1}$ or $a_n\geq a_{n+1}$, respectively.

Let's see if it is increasing. We want to verify if $\dfrac{8n}{1-2n}\leq\dfrac{8(n+1)}{1-(2n+1)}$

By simplifying the factor $8$ and multiplying by the denominators, while observing that the denominators are always negative, we obtain $$n(1-2(n+1))\leq (n+1)(1-2n)$$ By expanding the products we have $$-n-2n^2\leq +1-2n^2-n$$

And by reducing the term $-n-2n^2$ we obtain $0\leq1$ which is always true independently of $n$. Therefore the sequence is increasing.

To see if the sequence is definitely increasing we must verify whether $a_n < a_{n+1}$.

We can verify it using the previous calculations about the solution being definitely increasing since the calculations that were carried out are true if we replace the inequality $\leq$ by the strict inequality $ < $.

We verify if the sequence admits any bounds. As the sequence is increasing it is lower bounded by $a_1=-8$.

To see if the sequence is upper bounded we can see that $\dfrac{8n}{1-2n} < 0$ since the numerator is positive and the denominator is negative.

Therefore the sequence is upper bounded by $0$.

b) The sequence is not constant since $b_1=1$ and $b_2=\dfrac{4}{5}$.

Taking a look at the first two terms ot the sequence, it cannot be an increasing one. Let's see if the sequence is strictly decreasing. We verify if $$\dfrac{2n}{1+n^2} > \dfrac{2(n+1)}{1+(n+1)^2}$$ Multiplying by the denominators; $$2n(1+(n+1)^2) > (1+n^2)2(n+1)$$ Expanding we obtain; $$4n+4n^2+2n^3 > 2+2n+2n^2+2n^3$$ Simplifying we obtain $$n^2+n-1 > 0$$ Computing two roots of the previous polynomial we see that they are both smaller than $1$. Therefore, for $n$ integer it is satisfied that $n^2+n-1 > 0$ and the sequence is definitely decreasing.

As we have already seen earlier, in this case the sequence is upper bounded by $b_1=1$.

Also, since all the terms of the sequence are positive we establish that the sequence is lower bounded by $0$.

c) The sequence is not constant since $c_1=-\dfrac{3}{2}$ and $c_2=-2$.

By taking a look at the first two terms it can happen that the sequence is definitely decreasing. We verify if $$\dfrac{n^2+2}{-n-1} > \dfrac{(n+1)^2+2}{-(n+1)-1}$$ By multiplying by the denominators we obtain $$(-n-2)(n^2+2) > (n^2+2n+3)(-n-1)$$ By multiplying by $-1$, and therefore inverting the inequality and expanding, we obtain; $$n^3+2n^2+2n+4 < n^3+3n^2+5n+3$$ By subtracting we obtain the inequality $$n^2+3n-1 > 0$$

By calculating the roots we see that the two of them are smaller than 1 and, therefore, the inequality is true for every integer n.

Therefore, the sequence is strictly decreasing.

Consequently, the sequence is upper bounded by $c_1=-\dfrac{3}{2}$.

The sequence does not have a lower bound since the general term of the sequence becomes as big, with a negative sign, as desired.