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

#### wishmaster

##### Active member
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
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!
$$a_n = \frac{n^2 +1}{2n^2}$$

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

can we find "n" such that

$$\frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81}$$ ?
solve it for n

#### wishmaster

##### Active member
Re: Cant solve a sequence

$$a_n = \frac{n^2 +1}{2n^2}$$

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

can we find "n" such that

$$\frac{1}{2} + \frac{1}{2n^2} = \frac{41}{81}$$ ?
solve it for n
thank you,but i still dont get it.......

#### MarkFL

Staff member
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
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

Staff member
Re: Cant solve a sequence

Can 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?

#### wishmaster

##### Active member
Re: Cant solve a sequence

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?
And again,thank you!

#### MarkFL

Staff member
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
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

Staff member
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
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

Staff member
Re: Cant solve a sequence

81 = n2 ??
Yes, good!

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

#### wishmaster

##### Active member
Re: Cant solve a sequence

Yes, good!

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

Staff member
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
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

Staff member
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.

#### wishmaster

##### Active member
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.

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

#### MarkFL

Staff member
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
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

Staff member
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
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

Staff member
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
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

Staff member
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
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?

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