Limit of a sequence

Jan 2010
142
0
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....
 
Nov 2009
517
130
Big Red, NY
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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Jan 2010
142
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\(\displaystyle n\) 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

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Nov 2009
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Berkeley, California
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?
 
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Jan 2010
142
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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)
 
Nov 2009
517
130
Big Red, NY
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?
 
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Jan 2010
142
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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?
 
Nov 2009
517
130
Big Red, NY
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.
 
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