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#### wishmaster

##### Active member

- Oct 11, 2013

- 211

So the sequence is:

a

_{n}= \(\displaystyle \frac{2n+3}{n}\)

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- Thread starter
- #1

- Oct 11, 2013

- 211

So the sequence is:

a

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- #3

- Oct 11, 2013

- 211

I would say that upper limit is 5 and lower limit is 2.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?

But how to prove it? Or show it in math way?

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- #5

- Oct 11, 2013

- 211

I have assumed for aHow did you determine the bounds?

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You know:I have assumed for a_{1}

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

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?

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- #7

- Oct 11, 2013

- 211

\(\displaystyle \lim _{n \to \infty}2+ \frac{3}{n} = 2 + 0=0\) ???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?

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

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- #9

- Oct 11, 2013

- 211

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.....

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- #10

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: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.....

\(\displaystyle \frac{3}{n+1}-\frac{3}{n}\)

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