# Problems from Bounded sequences

Determine the behavior of the following sequences and calculate if there is a possible upper and lower bound.

a) $$a_n=\dfrac{8n}{1-2n}$$

b) $$b_n=\dfrac{2n}{1+n^2}$$

c) $$c_n=\dfrac{n^2+2}{-n-1}$$

See development and solution

### Development:

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.

### Solution:

a) The sequence is strictly increasing. It is upper bounded by $$0$$ and lower bounded by $$-8$$.

b) The sequence is strictly decreasing. It is upper bounded by $$1$$ and lower bounded by $$0$$.

c) The sequence is strictly decreasing. It is upper bounded by $$-\dfrac{3}{2}$$ and does not admit any lower bound.

Hide solution and development