Proving a sequence is monotone.

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Homework Help Overview

The discussion revolves around proving that the sequence defined by √n + 1/n is monotone increasing, specifically that a_n+1 > a_n for n greater than 2. The original poster has established that the sequence is bounded from below but not from above and seeks assistance in demonstrating the monotonicity of the sequence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods to prove the inequality a_n+1 > a_n, including the possibility of using induction and the need to express the terms explicitly. There is also a focus on algebraic manipulation to demonstrate the positivity of the difference between consecutive terms.

Discussion Status

The discussion is ongoing, with participants providing suggestions and exploring different angles to approach the problem. Some guidance has been offered regarding the algebraic expression of the terms, but there is no consensus on a definitive method to prove the monotonicity yet.

Contextual Notes

Participants note that the proof must be conducted without reference to functions, which adds a constraint to the approaches being considered. The original poster has confirmed that the inequality holds for n=2, but further algebraic demonstration is required for larger n.

peripatein
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Hi,

I am unsuccessful at showing that the sequence √n + 1/n is an ascending monotone one, i.e. that a_n+1 > a_n for any n, greater than 2 let's say. I have proven that it is not bounded from above and is bounded from below. Any ideas, suggestions, please?
 
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peripatein said:
Hi,

I am unsuccessful at showing that the sequence √n + 1/n is an ascending monotone one, i.e. that a_n+1 > a_n for any n, greater than 2 let's say. I have proven that it is not bounded from above and is bounded from below. Any ideas, suggestions, please?
This appears to be essentially the same as a previous thread you started:

https://www.physicsforums.com/showthread.php?t=649117
 
Hi,
Except that no one there was able to help me. Perhaps someone could now.
 
Consider your sequence as a function.
 
I am not allowed to. It has to be proven without any reference to functions. Seemingly merely by showing that a_n+1 > a_n for any n greater than 2. Yet I am unable to show that that inequality holds.
 
Write out what a_(n+1) actually is.
 
sqrt(n+1) + 1/(n+1).
 
Now what do you do to prove one sequence is larger than the other for all n≥2.
 
Induction? I have been unsuccessful at proving it via induction as well.
 
  • #10
If [itex]a_{n+1} > a_n \Rightarrow a_{n+1} - a_n > 0[/itex]
 
  • #11
And?
 
  • #12
Write out what it means.
 
  • #13
sqrt(n+1) + 1/(n+1) - sqrt(n) - 1/n > 0
 
  • #14
peripatein said:
sqrt(n+1) + 1/(n+1) - sqrt(n) - 1/n > 0

You could clean it up a bit, but is that sequence greater than zero for every n?

If it helps, roughly around n = 1.2, your sequence geometrically will start to be above the line y=0.
 
Last edited:
  • #15
It surely is greater than zero but I am unable to demonstrate it algebrically, alas.
 
  • #16
peripatein said:
It surely is greater than zero but I am unable to demonstrate it algebrically, alas.

So, start with n≥2.

For n=2 does the inequality hold?
 
  • #17
It does.
 

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