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wishmaster
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- Oct 11, 2013
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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]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!
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
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.
I would rather help you continue, so that you learn more.Can you continiue?
And many thanks
Yes,i understand this!I would rather help you continue, 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?
81n2+1 = 82n2You 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.
When you distribute the $81$ on the left side, you need to do so to both terms within the parentheses, as follows: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?
Yes, good!81 = n2 ??
9 of course.Yes, good!
So, what is the positive root? What is the positive value of $n$?
Yes, you have shown that:9 of course.
But what have i proved?
That 41/81 is a 9th element of sequence?
Ok,thank you.Yes, you have shown that:
\(\displaystyle a_{9}=\frac{41}{81}\)
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.![]()
No, that's not the case at all...let's look at the $n$th term:Seems im to stupid for sequences......
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?
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......
1/2 ?What does \(\displaystyle \frac{1}{2n^2}\) approach as $n$ grows without bound?
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
I think that bn would be zero,or very close.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?
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?