Exact upper and lower limit of the sequence

[theakdad]

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 don't know how to do it.

So the sequence is:

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

 

[MarkFL]

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

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

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

 

[theakdad]

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

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

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]

How did you determine the bounds?

 

[theakdad]

How did you determine the bounds?

I have assumed for a₁

 

[MarkFL]

I have assumed for a₁

You know:

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

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

 

[theakdad]

You know:

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

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

$$\lim _{n \to \infty}2+ \frac{3}{n} = 2 + 0=0$$ ?

 

[MarkFL]

$$\lim _{n \to \infty}2+ \frac{3}{n} = 2 + 0=0$$ ?

Correct, except that $2+0=2$.

 

[theakdad]

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]

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...

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:

$$\frac{3}{n+1}-\frac{3}{n}$$

And show that it is negative for all $n\in\mathbb{N}$.