# Exact upper and lower limit of the sequence

#### wishmaster

##### Active member
I have one more problem,for the given sequence i have to find the exact upper and lower limit,and to argument them. i have been missing on this lesson,so please help me,i dont know how to do it.

So the sequence is:

an = $$\displaystyle \frac{2n+3}{n}$$

#### MarkFL

Staff member
If I am interpreting correctly what you meant by upper and lower limit, I would rewrite the $n$th term as follows:

$$\displaystyle a_n=2+\frac{3}{n}$$

What can you say about the terms as $n$ grows?

#### wishmaster

##### Active member
If I am interpreting correctly what you meant by upper and lower limit, I would rewrite the $n$th term as follows:

$$\displaystyle a_n=2+\frac{3}{n}$$

What can you say about the terms as $n$ grows?
I would say that upper limit is 5 and lower limit is 2.
But how to prove it? Or show it in math way?

#### MarkFL

Staff member
How did you determine the bounds?

#### MarkFL

Staff member
I have assumed for a1
You know:

$$\displaystyle a_1=2+\frac{3}{1}=5$$

and you know that $$\displaystyle \frac{3}{n}$$ gets smaller as $n$ increases, so then you know $a_n$ is monotonically decreasing. How can you determine the lower bound?

#### wishmaster

##### Active member
You know:

$$\displaystyle a_1=2+\frac{3}{1}=5$$

and you know that $$\displaystyle \frac{3}{n}$$ gets smaller as $n$ increases, so then you know $a_n$ is monotonically decreasing. How can you determine the lower bound?
$$\displaystyle \lim _{n \to \infty}2+ \frac{3}{n} = 2 + 0=0$$ ???

#### MarkFL

Staff member
$$\displaystyle \lim _{n \to \infty}2+ \frac{3}{n} = 2 + 0=0$$ ???
Correct, except that $2+0=2$.

#### wishmaster

##### Active member
Correct, except that $2+0=2$.

Im sorry,thats what i thought..my mistake.

So with limit i have proved exact lower limit,i know that exact upper limit is 5,but how to write it correctly? With proof.....

#### MarkFL

What more do you need? You know the sequence bounds and that it is monotonic. All you need to do is discuss the fact that $\dfrac{3}{n}$ decreases as $n$ increases. Or you could consider the difference:
$$\displaystyle \frac{3}{n+1}-\frac{3}{n}$$
And show that it is negative for all $n\in\mathbb{N}$.