Can't solve a sequence (to determine if a given value is a member)

[theakdad]

Can somebody help me with this problem? I have to proof if 41/81 is a part of attached sequence. Step by step guide would be useful. Thank you! http://img.tapatalk.com/d/13/10/18/2unyje8y.jpg

 

[Amer]

Re: Cant solve a sequence

Can somebody help me with this problem? I have to proof if 41/81 is a part of attached sequence. Step by step guide would be useful. Thank you!

[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

 

[theakdad]

Re: Cant solve a sequence

[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

thank you,but i still don't get it...

 

[MarkFL]

Re: Cant solve a sequence

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.

 

[theakdad]

Re: Cant solve a sequence

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.

Can you continiue?
And many thanks

 

[MarkFL]

Re: Cant solve a sequence

Can you continiue?
And many thanks

I would rather help you continue, so that you learn more. :D

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?

 

[theakdad]

Re: Cant solve a sequence

I would rather help you continue, so that you learn more. :D

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?

Yes,i understand this!

So i think that's 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!

 

[MarkFL]

Re: Cant solve a sequence

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.

 

[theakdad]

Re: Cant solve a sequence

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.

81n²+1 = 82n²
1 = 82n²-81nⁿ
1= n²

I don't know i I am right...or what to do...

 

[MarkFL]

Re: Cant solve a sequence

81n²+1 = 82n²
1 = 82n²-81nⁿ
1= n²

I don't know i I am right...or what to do...

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?

 

[theakdad]

Re: Cant solve a sequence

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 = n² ??

 

[MarkFL]

Re: Cant solve a sequence

81 = n² ??

Yes, good! :D

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

 

[theakdad]

Re: Cant solve a sequence

Yes, good! :D

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

9 of course.

But what have i proved?That 41/81 is a 9th element of sequence?

 

[MarkFL]

Re: Cant solve a sequence

9 of course.

But what have i proved?That 41/81 is a 9th element of sequence?

Yes, you have shown that:

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

 

[theakdad]

Re: Cant solve a sequence

Yes, you have shown that:

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

Ok,thank you.

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...

 

[MarkFL]

Without using calculus, to explore the monotonicity of the sequence, I would write the $n$th term as suggested above by 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. :D

 

[theakdad]

Without using calculus, to explore the monotonicity of the sequence, I would write the $n$th term as suggested above by 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. :D

Seems I am to stupid for sequences...

 

[MarkFL]

Seems I am to stupid for sequences...

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?

 

[theakdad]

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?

Yes,i agree!
I think that this sequence is limited too 0,5, but i don't 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 don't know how would i master this all things...

 

[MarkFL]

Yes,i agree!
I think that this sequence is limited too 0,5, but i don't 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 don't know how would i master this all things...

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

 

[theakdad]

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

1/2 ?

I really don't know how to get it...

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

 

[MarkFL]

1/2 ?

I really don't know how to get it...

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

Let'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?

 

[theakdad]

Let'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?

I think that b_n would be zero,or very close.
but beacuse of + 1/2 b_n can't be smaler then 0,5. Am i right?

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

 

[MarkFL]

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?

 

[theakdad]

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?

Yes,it makes sense ,of course. Only problem si,how to write this in a proper way for my profesor? :pAnd why is \(\displaystyle \frac{n^2+1}{2n^2}\) equal to \(\displaystyle \frac{1}{2} + \frac{1}{2n^2}\) ?

 

[MarkFL]

Yes,it makes sense ,of course. Only problem si,how to write this in a proper way for my profesor? :pAnd why is \(\displaystyle \frac{n^2+1}{2n^2}\) equal to \(\displaystyle \frac{1}{2} + \frac{1}{2n^2}\) ?

A property of fractions is that:

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

so that:

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

 

[theakdad]

A property of fractions is that:

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

so that:

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

Now i see...and understand.

Only thing that's left,how to write it on the paper,so that would be acteptable for my teacher...

I owe you a drink now :D

 

[MarkFL]

From what course is this problem? If it is for a course prior to calculus, then you will simply need to explain it as best you can in your own words. You could also choose to write the $n$th term as:

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

Now it is easy to see that the denominator is fixed, and the numerator is getting smaller and smaller as $n$ gets larger and larger, that is, the sequence is strictly decreasing, i.e., it is monotonic.

 

[theakdad]

From what course is this problem? If it is for a course prior to calculus, then you will simply need to explain it as best you can in your own words. You could also choose to write the $n$th term as:

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

Now it is easy to see that the denominator is fixed, and the numerator is getting smaller and smaller as $n$ gets larger and larger, that is, the sequence is strictly decreasing, i.e., it is monotonic.

Its prior to algebra...

I will try to do somehow...

 

[MarkFL]

I would think the argument I gave in my post previous to this would suffice then.

 

[theakdad]

I would think the argument I gave in my post previous to this would suffice then.

I agree...

But,now i have new problem,about Mathematical induction,should i open new Thread?

 

[MarkFL]

I agree...

But,now i have new problem,about Mathematical induction,should i open new Thread?

Yes, and I would post it in our Discrete Mathematics sub-forum.

 

[theakdad]

Yes, and I would post it in our Discrete Mathematics sub-forum.

I will...

Maybe you can take a look...if you want...

 

[theakdad]

I still have one question...What kind of type this sequence is? Arithmetic or geometric? I can't figure it out...

 

[MarkFL]

I still have one question...What kind of type this sequence is? Arithmetic or geometric? I can't figure it out...

An arithmetic sequence has the property:

\(\displaystyle a_{n+1}-a_{n}=d\)

While a geometric sequence would have the property:

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

Note: Both $d$ and $r$ are constants that do not depend on $n$. Does this sequence satisfy either property?