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

Click For Summary
The discussion revolves around determining if the fraction 41/81 is a member of the sequence defined by a_n = (n^2 + 1) / (2n^2). Participants guide the original poster through the process of cross-multiplying and simplifying the equation to find a natural number solution for n. After deriving the equation 81(n^2 + 1) = 82n^2, they confirm that n equals 9, indicating that 41/81 is indeed the 9th element of the sequence. The conversation also touches on exploring the monotonicity of the sequence and its graphical representation, as well as distinguishing between arithmetic and geometric sequences. Ultimately, the original poster expresses gratitude for the assistance and seeks further clarification on related mathematical concepts.
  • #31
MarkFL said:
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?
 
Mathematics news on Phys.org
  • #32
andreask said:
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.
 
  • #33
MarkFL said:
Yes, and I would post it in our Discrete Mathematics sub-forum.

I will...

Maybe you can take a look...if you want...
 
  • #34
I still have one question...What kind of type this sequence is? Arithmetic or geometric? I can't figure it out...
 
  • #35
andreask said:
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:

$$a_{n+1}-a_{n}=d$$

While a geometric sequence would have the property:

$$\frac{a_{n+1}}{a_{n}}=r$$

Note: Both $d$ and $r$ are constants that do not depend on $n$. Does this sequence satisfy either property?
 
  • #36
MarkFL said:
An arithmetic sequence has the property:

$$a_{n+1}-a_{n}=d$$

While a geometric sequence would have the property:

$$\frac{a_{n+1}}{a_{n}}=r$$

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

It seems its arthmetic
 
  • #37
andreask said:
It seems its arthmetic

I find:

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

Thus, the difference between two succeeding terms is a function of $n$, and so the sequence is not arithmetic.
 
  • #38
MarkFL said:
I find:

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

Thus, the difference between two succeeding terms is a function of $n$, and so the sequence is not arithmetic.

Ad its not geometric?
 
  • #39
andreask said:
And its not geometric?

What do you find when you compute the ratio I gave above?
 
  • #40
MarkFL said:
What do you find when you compute the ratio I gave above?

Will try tomorrow,im litlle busy now
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K