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

wishmaster

Active member
Oct 11, 2013
211
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!
 

Amer

Active member
Mar 1, 2012
275
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
 

wishmaster

Active member
Oct 11, 2013
211
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 dont get it.......
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

wishmaster

Active member
Oct 11, 2013
211
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 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!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

wishmaster

Active member
Oct 11, 2013
211
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 dont know i im right.....or what to do.....
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Cant solve a sequence

81n2+1 = 82n2
1 = 82n2-81nn
1= n2

I dont know i im 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?
 

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
 

wishmaster

Active member
Oct 11, 2013
211
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 im to stupid for sequences......
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Seems im 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?
 

wishmaster

Active member
Oct 11, 2013
211
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 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......
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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......
What does \(\displaystyle \frac{1}{2n^2}\) approach as $n$ grows without bound?
 

wishmaster

Active member
Oct 11, 2013
211
What does \(\displaystyle \frac{1}{2n^2}\) approach as $n$ grows without bound?
1/2 ?

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

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
1/2 ?

I really dont 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?
 

wishmaster

Active member
Oct 11, 2013
211
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 bn would be zero,or very close.
but beacuse of + 1/2 bn cant be smaler then 0,5. Am i right?

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

wishmaster

Active member
Oct 11, 2013
211
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? :p


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