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:
$\displaystyle a_n = \frac{2n+1}{n+1} > 0$
 A: An infinite sequence $a_1, a_2, a_3, \dots$ of real numbers is said to be bounded above if there is a number $U$ such that for every $n$, $a_n \le U$.  This is a slightly fancy way of saying that there is a "ceiling" $U$ such that every member of the sequence is $\le U$. Any such ceiling is called an upper bound of the sequence.
The sequence is bounded below if there is a number $L$ such that every member of the sequence is $\ge L$. So the sequence is bounded below if there is a "floor" $L$ below which the elements of the sequence can't go. Any such $L$ is called a lower bound of the sequence. 
We say that the sequence is bounded if it is both bounded above and bounded below.
So to prove that a particular sequence is bounded, you need to show two things, that it is bounded above and that it is bounded below.  Most of the time, it is unwise to try to prove two different things in one calculation. So the proof will consist of two separate parts.
For your particular sequence, proving the "bounded below" part is quite easy.  We want to come up with a number $L$ such that $L \le a_n$, a "floor" $L$ below which you can guarantee that there can't be any $a_n$.
It is clear that every member of your sequence is positive.  So you can guarantee that $0 \le a_n$ for every $n$.  Thus $0$ is a lower bound fof the sequence, and it follows that the sequence is bounded below.
Note that $-17$ is also a lower bond for the sequence.  In order to prove that a sequence is bounded below, you need not come up with a "best" lower bound, any lower bound will do.
Now let's tackle the upper bound problem.  You might wish to experiment with a calculator.  So calculate $\frac{2n+1}{n+1}$ for various values of $n$.  You will observe that the numbers you calculate are all below $2$.  Could it be that $2$ is an upper bound ("ceiling") for the sequence?
The proof that it is has been described by others.  You can see that the numerator is always less than $2n+2$, which means that
$$a_n <\frac{2n+2}{n+1}=2$$
So $a_n \le 2$ for all $n$, and therefore $2$ is an upper bound ("ceiling") for our sequence.  The above calculation is a bit too slick, and maybe hard to come up with until you get some experience, so I will give a less slick, but equally correct, approach below.
It is interesting to observe, either from calculator experimentations or by using some algebra, that for large $n$, $a_n$ is (of course) below $2$, but awfully close to $2$.  It turns out that $2$ is the cheapest possible ceiling for your sequence, no number below $2$ canbe a ceiling for your sequence.
But remember that to prove that a sequence is bounded, all you need to do is to come up with some upper bound.  Let's show that $17$ is an upper bound for our sequence.
So we want to prove that 
$$\frac{2n+1}{n+1} \le 17$$
for all $n$.  By basic properties of inequalities, this is equivalent to showing that $2n+1 \le 17(n+1)$
for all $n$.  And that is equivalent to showing that for all $n$, $2n+1<17n+17$, which is of course obvious.  So you have shown that $17$  is an upper bound for the sequence, and therefore the sequence is bounded. The algebra would in fact work out the same way if instead of $17$ you used any number $\ge 2$.  Try it with $2$, to get your own proof that in fact $2$ is an upper bound.  
A: $\langle a_{n} \rangle$ is a sequence of rationals.
You can think of it as a function that associates with each natural number, $n$, a rational number, in this case $\frac{2n + 1}{n + 1}$.
To be bounded means that there are numbers (bounds) beyond which the sequence (i.e. the value of the function) cannot go, in this case, 2 and 0.
What the key proves is that for each $n$, $a_{n}$ cannot be greater than $2$ and cannot be smaller than $0$. Thus it doesn't matter how far down the sequence you go, the sequence will always be between these two bounds.
A simple way to visualise this is to draw the graph of $a_{n}$ and draw the two straight lines $y = 0$ and $y = 2$.  You will then see that the graph of $a_{n}$ never steps outside these two lines.
A: 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.

