<s>[INST] How to prove that $a_n = \frac{2n+1}{n+1}$ is bounded? How do I prove that $\displaystyle a_n = \frac{2n+1}{n+1}$ is bounded?
According to the key it works as follows. However, I don't really understand the meaning.
Upper bound:
$\displaystyle a_n = \frac{2n+1}{n+1} < \frac{2n+2}{n+1} = \frac{2(n+1)}{n+1} = 2$
Lower bound:
$\displaystyle a_n = \frac{2n+1}{n+1} > 0$
 [/INST] For the upper bound part you could also reason as follows:
$$\frac{2n+1}{n+1}=\frac{n+n+1}{n+1}=\frac{n}{n+1}+\frac{n+1}{n+1}=\frac{n}{n+1}+1$$
Since $\dfrac n {n+1}$ comes as monotonically increasing on the natural numbers, and the limit of  $\dfrac n {n+1}$ as $n$ increases without bound equals 1, it follows that the limit of $\dfrac{2n+1}{n+1}$ as $n$ increases without bound equals $2$.  Since this limit happens for a monotonically increasing function, it qualifies as a least upper bound.  So, we have an upper bound.
One could also precede as follows:


*

*$0<1$

*$(2n+1)+0<(2n+1)+1$ by monotonicity of $+$

*$2n+1<(2n+1)+1$ by $0$ as the neutral for $+$

*$2n+1<2n+(1+1)$ by association of $+$

*$2n+1<2n+2$  by $1+1=2$

*$2n+1<2(n+1)$ by distribution of multiplication over addition, and $2*1=2$

*$\dfrac{2n+1}{n+1}<2$ several steps here actually, and it only works since $n+1>0$ by definition.

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