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#### wishmaster

##### Active member

- Oct 11, 2013

- 211

- Thread starter wishmaster
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- #1

- Oct 11, 2013

- 211

[tex] a_n = \frac{n^2 +1}{2n^2} [/tex]

[tex]\frac{n^2 +1}{2n^2} = \frac{n^2}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{2n^2} [/tex]

can we find "n" such that

[tex] \frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81} [/tex] ?

solve it for n

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- #3

- Oct 11, 2013

- 211

thank you,but i still dont get it.......[tex] a_n = \frac{n^2 +1}{2n^2} [/tex]

[tex]\frac{n^2 +1}{2n^2} = \frac{n^2}{2n^2} + \frac{1}{2n^2} = \frac{1}{2} + \frac{1}{2n^2} [/tex]

can we find "n" such that

[tex] \frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81} [/tex] ?

solve it for n

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- #5

- Oct 11, 2013

- 211

Can you continiue?I would write:

\(\displaystyle \frac{n^2+1}{2n^2}=\frac{41}{81}\)

Now, cross-multiply and then see if you can find a natural number solution.

And many thanks

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- #6

I would rather helpCan you continiue?

And many thanks

What do you get when you cross-multiply? If you are unfamiliar with this method, if you have:

\(\displaystyle \frac{a}{b}=\frac{c}{d}\)

then cross-multiplying, which means to multiply the numerators on each side by the denominators on the other, and then equating both products, will give you:

\(\displaystyle ad=bc\)

It is a kind of shortcut for multiplying both sides by the common denominator of $bd$:

\(\displaystyle \frac{a}{b}bd=\frac{c}{d}bd\)

Canceling, or reducing the fractions, we get:

\(\displaystyle ad=bc\)

Can you do this with the equation I gave?

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- #7

- Oct 11, 2013

- 211

Yes,i understand this!I would rather helpyoucontinue, so that you learn more.

What do you get when you cross-multiply? If you are unfamiliar with this method, if you have:

\(\displaystyle \frac{a}{b}=\frac{c}{d}\)

then cross-multiplying, which means to multiply the numerators on each side by the denominators on the other, and then equating both products, will give you:

\(\displaystyle ad=bc\)

If it a kind of shortcut for multiplying both sides by the common denominator of bd:

\(\displaystyle \frac{a}{b}bd=\frac{c}{d}bd\)

Canceling, or reducing the fractions, we get:

\(\displaystyle ad=bc\)

Can you do this with the equation I gave?

So i think thats my result: 81*(n^2+1) = 82(n^2)

So 41/81 is not a member of my sequence. Am i right?

I would love to learn more about such sequences,do you know some good reference?

And again,thank you!

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- #8

You have the correct equation:

\(\displaystyle 81\left(n^2+1 \right)=41\left(2n^2 \right)\)

Now, distribute on both sides, then see if you can solve for $n$...

And no, I do not know of any online resources for studying sequences specifically.

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- #9

- Oct 11, 2013

- 211

81nYou have the correct equation:

\(\displaystyle 81\left(n^2+1 \right)=41\left(2n^2 \right)\)

Now, distribute on both sides, then see if you can solve for $n$...

And no, I do not know of any online resources for studying sequences specifically.

1 = 82n

1= n

I dont know i im right.....or what to do.....

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- #10

When you distribute the $81$ on the left side, you need to do so to both terms within the parentheses, as follows:81n^{2}+1 = 82n^{2}

1 = 82n^{2}-81n^{n}

1= n^{2}

I dont know i im right.....or what to do.....

\(\displaystyle 81n^2+81=82n^2\)

Next, subtract $81n^2$ from both sides, and what are we left with?

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- #11

- Oct 11, 2013

- 211

When you distribute the $81$ on the left side, you need to do so to both terms within the parentheses, as follows:

\(\displaystyle 81n^2+81=82n^2\)

Next, subtract $81n^2$ from both sides, and what are we left with?

81 = n

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- #12

Yes, good!81 = n^{2}??

So, what is the positive root? What is the positive value of $n$?

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- #13

- Oct 11, 2013

- 211

9 of course.Yes, good!

So, what is the positive root? What is the positive value of $n$?

But what have i proved?

That 41/81 is a 9th element of sequence?

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- #14

Yes, you have shown that:9 of course.

But what have i proved?

That 41/81 is a 9th element of sequence?

\(\displaystyle a_{9}=\frac{41}{81}\)

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- #15

- Oct 11, 2013

- 211

Ok,thank you.Yes, you have shown that:

\(\displaystyle a_{9}=\frac{41}{81}\)

i still have three questions:

1. How can i explore the monotonoty of this sequence?

2. How to draw this sequence in coordinate system?

3. Can you show me another sequence solution(similar to mine)? Of course,if you have time.....

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- #16

\(\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}\)

Can you see that each succeeding term is smaller than the last, and that as $n$ grows without bound $a_{n}$ asymptotically approaches \(\displaystyle \frac{1}{2}\)?

You could plot the first few terms, along with the asymptote.

You could make up another such problem just as easily as I can.

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- #17

- Oct 11, 2013

- 211

Amer:

\(\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}\)

Can you see that each succeeding term is smaller than the last, and that as $n$ grows without bound $a_{n}$ asymptotically approaches \(\displaystyle \frac{1}{2}\)?

You could plot the first few terms, along with the asymptote.

You could make up another such problem just as easily as I can.

Seems im to stupid for sequences......

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- #18

No, that's not the case at all...let's look at the $n$th term:Seems im to stupid for sequences......

\(\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}\)

And let's for now just look at this part of the $n$th term:

\(\displaystyle \frac{1}{2n^2}\)

Would you agree that when the numerator of a fraction is fixed, that is remains constant, and the denominator gets bigger and bigger, then the value of the fraction is getting smaller and smaller?

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- #19

- Oct 11, 2013

- 211

Yes,i agree!No, that's not the case at all...let's look at the $n$th term:

\(\displaystyle a_{n}=\frac{1}{2}+\frac{1}{2n^2}\)

And let's for now just look at this part of the $n$th term:

\(\displaystyle \frac{1}{2n^2}\)

Would you agree that when the numerator of a fraction is fixed, that is remains constant, and the denominator gets bigger and bigger, then the value of the fraction is getting smaller and smaller?

I think that this sequence is limited too 0,5, but i dont know how to prove it.....

i have started a master degree with 38,my math was very good at high school,but now...now is all at higher level,i really dont know how would i master this all things......

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- #20

What does \(\displaystyle \frac{1}{2n^2}\) approach as $n$ grows without bound?Yes,i agree!

I think that this sequence is limited too 0,5, but i dont know how to prove it.....

i have started a master degree with 38,my math was very good at high school,but now...now is all at higher level,i really dont know how would i master this all things......

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- #21

- Oct 11, 2013

- 211

1/2 ?What does \(\displaystyle \frac{1}{2n^2}\) approach as $n$ grows without bound?

I really dont know how to get it......

And i really apreciate that you have time for me,thank you

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- #22

Let's look at the first few term of this fraction...let:1/2 ?

I really dont know how to get it......

And i really apreciate that you have time for me,thank you

\(\displaystyle b_{n}=\frac{1}{2n^2}\)

We find:

\(\displaystyle b_1=\frac{1}{2(1)^2}=\frac{1}{2}\)

\(\displaystyle b_2=\frac{1}{2(2)^2}=\frac{1}{8}\)

\(\displaystyle b_3=\frac{1}{2(3)^2}=\frac{1}{18}\)

\(\displaystyle b_4=\frac{1}{2(4)^2}=\frac{1}{32}\)

\(\displaystyle \vdots\)

\(\displaystyle b_{1000}=\frac{1}{2(1000)^2}=\frac{1}{2000000}\)

\(\displaystyle \vdots\)

\(\displaystyle b_{1000000}=\frac{1}{2(1000000)^2}=\frac{1}{2000000000000}\)

Do you see that as $n$ gets bigger and bigger, $b_{n}$ gets smaller and smaller? If we could let $n$ go to infinity, then what would $b_{n}$ be?

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- #23

- Oct 11, 2013

- 211

I think that bLet's look at the first few term of this fraction...let:

\(\displaystyle b_{n}=\frac{1}{2n^2}\)

We find:

\(\displaystyle b_1=\frac{1}{2(1)^2}=\frac{1}{2}\)

\(\displaystyle b_2=\frac{1}{2(2)^2}=\frac{1}{8}\)

\(\displaystyle b_3=\frac{1}{2(3)^2}=\frac{1}{18}\)

\(\displaystyle b_4=\frac{1}{2(4)^2}=\frac{1}{32}\)

\(\displaystyle \vdots\)

\(\displaystyle b_{1000}=\frac{1}{2(1000)^2}=\frac{1}{2000000}\)

\(\displaystyle \vdots\)

\(\displaystyle b_{1000000}=\frac{1}{2(1000000)^2}=\frac{1}{2000000000000}\)

Do you see that as $n$ gets bigger and bigger, $b_{n}$ gets smaller and smaller? If we could let $n$ go to infinity, then what would $b_{n}$ be?

but beacuse of + 1/2 b

And sorry,i have to learn to write fractions etc. in right form.

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- #25

- Oct 11, 2013

- 211

Yes,it makes sense ,of course. Only problem si,how to write this in a proper way for my profesor?Yes, $b_{n}$ gets closer and closer to zero, and since:

\(\displaystyle a_n=\frac{1}{2}+b_n\)

then \(\displaystyle a_{n}\) gets closer and closer to \(\displaystyle \frac{1}{2}\).

Does this make sense?

And why is \(\displaystyle \frac{n^2+1}{2n^2}\) equal to \(\displaystyle \frac{1}{2} + \frac{1}{2n^2}\) ?