# Limit of a sequence

#### Anemori

Forgive me, if this is the right forum.

Problem:

Convince me $$\displaystyle an=\frac{1}{n^2+n}$$ converge to zero.

I tried it with an=1

$$\displaystyle \frac{1}{1^2+1}=\frac{1}{2}$$

$$\displaystyle \frac{1}{(1/2)^2+(1/2)}=\frac{4}{3}$$

$$\displaystyle \frac{1}{(4/3)^2+(4/3)}=\frac{9}{28}$$

It doesnt go down to zero....

#### Anonymous1

Forgive me, if this is the right forum.

Problem:

Convince me $$\displaystyle an=\frac{1}{n^2+n}$$ converge to zero.

I tried it with an=1

$$\displaystyle \frac{1}{1^2+1}=\frac{1}{2}$$

$$\displaystyle \frac{1}{(1/2)^2+(1/2)}=\frac{4}{3}$$

$$\displaystyle \frac{1}{(4/3)^2+(4/3)}=\frac{9}{28}$$

It doesnt go down to zero....
$$\displaystyle n$$ is becoming $$\displaystyle \infty$$ly large, try plugging in some numbers like $$\displaystyle n=1000$$ and see what you get.

Here is one way to show this:

Convince yourself that, $$\displaystyle \frac{1}{n}$$ is becoming infinitely small as $$\displaystyle n\rightarrow \infty.$$ Likewise $$\displaystyle \frac{1}{n^2}$$ is becoming infinitely small as $$\displaystyle n\rightarrow \infty.$$

It follows that $$\displaystyle \frac{1}{n}\rightarrow 0$$ and $$\displaystyle \frac{1}{n^2}\rightarrow 0$$ as $$\displaystyle n\rightarrow \infty.$$

Now, take your sequence and divide the top and bottom by $$\displaystyle n^2.$$

$$\displaystyle a_n=\frac{1}{n^2+n} = \frac{\frac{1}{n^2}}{1 + \frac{1}{n}} \rightarrow \frac{0}{1+0} = 0$$ as $$\displaystyle n\rightarrow \infty.$$

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• Anemori

#### Anemori

$$\displaystyle n$$ is becoming $$\displaystyle \infty$$ly large, try plugging in some numbers like $$\displaystyle n=1000$$ and see what you get.

Here is one way to show this:

Convince yourself that, $$\displaystyle \frac{1}{n}$$ is becoming infinitely small as $$\displaystyle n\rightarrow \infty.$$ Likewise $$\displaystyle \frac{1}{n^2}$$ is becoming infinitely small as $$\displaystyle n\rightarrow \infty.$$

It follows that $$\displaystyle \frac{1}{n}\rightarrow 0$$ and $$\displaystyle \frac{1}{n^2}\rightarrow 0$$ as $$\displaystyle n\rightarrow \infty.$$

Now, take your sequence and divide the top and bottom by $$\displaystyle n^2.$$

$$\displaystyle a_n=\frac{1}{n^2+n} = \frac{\frac{1}{n^2}}{1 + \frac{1}{n}} \rightarrow \frac{0}{1+0} = 0$$ as $$\displaystyle n\rightarrow \infty.$$

How can you say that:
$$\displaystyle a_n=\frac{1}{n^2+n} = \frac{\frac{1}{n^2}}{1 + \frac{1}{n}} \rightarrow \frac{0}{1+0} = 0$$ as $$\displaystyle n\rightarrow \infty.$$

when you plug in n=0 becomes:

$$\displaystyle \frac{\frac{1}{0}}{1+{\frac{1}{0}}} = 0$$

when $$\displaystyle \frac {1}{0}$$ is not equal to zero but it is impossible.

if I plug $$\displaystyle n\rightarrow \infty$$ i get an alternate sequence $$\displaystyle a_n\rightarrow \infty$$ and $$\displaystyle a_n\rightarrow 0$$

but i need to convince that $$\displaystyle a_n$$ converge to zero.

#### Drexel28

MHF Hall of Honor
How can you say that:
$$\displaystyle a_n=\frac{1}{n^2+n} = \frac{\frac{1}{n^2}}{1 + \frac{1}{n}} \rightarrow \frac{0}{1+0} = 0$$ as $$\displaystyle n\rightarrow \infty.$$

when you plug in n=0 becomes:

$$\displaystyle \frac{\frac{1}{0}}{1+{\frac{1}{0}}} = 0$$

when $$\displaystyle \frac {1}{0}$$ is not equal to zero but it is impossible.

if I plug $$\displaystyle n\rightarrow \infty$$ i get an alternate sequence $$\displaystyle a_n\rightarrow \infty$$ and $$\displaystyle a_n\rightarrow 0$$

but i need to convince that $$\displaystyle a_n$$ converge to zero.
Your sequence is less than $$\displaystyle \frac{1}{n}$$. Now, what does the Archimedean principle say?

• Anemori

#### Anemori

Your sequence is less than $$\displaystyle \frac{1}{n}$$. Now, what does the Archimedean principle say?

Honestly I don't know any archimedean principle.. (Crying)

#### Anonymous1

Honestly I don't know any archimedean principle.. (Crying)
I think there is a fundamental misunderstanding of sequences here.

You do not plug the answer you get back into the sequence to obtain the next term. Most likely, you are confusing this with recursion.

You are trying to show that as n becomes arbitrarily large, your sequence becomes arbitrarily small.

The sequence looks like this:

$$\displaystyle a_{(1)}, a_{(2)}, a_{(3)}, a_{(4)},...,a_{(1000)},...$$

$$\displaystyle \frac{1}{(1)^2 + (1)},\frac{1}{(2)^2 + (2)},...,\frac{1}{(1000)^2 + (1000)},...$$

Now, what is happening to the sequence?

• Anemori

#### Anemori

I think there is a fundamental misunderstanding of sequences here.

You do not plug the answer you get back into the sequence to obtain the next term. Most likely, you are confusing this with recursion.

You are trying to show that as n becomes arbitrarily large, your sequence becomes arbitrarily small.

The sequence looks like this:

$$\displaystyle a_{(1)}, a_{(2)}, a_{(3)}, a_{(4)},...,a_{(1000)},...$$

$$\displaystyle \frac{1}{(1)^2 + (1)},\frac{1}{(2)^2 + (2)},...,\frac{1}{(1000)^2 + (1000)},...$$

Now, what is happening to the sequence?
I see, then what is that sequence that i plug in the answer to get the next sequence to get the next term?

#### Anonymous1

I see, then what is that sequence that i plug in the answer to get the next sequence to get the next term?
I'm not sure I know what this means.

Everything you need is in my previous two posts. Go read them again, and carefully too.

• Anemori

#### Anemori

I'm not sure I know what this means.

Everything you need is in my previous two posts. Go read them again, and carefully too.
Thanks! (Clapping)