Proving a sequence is monotone.

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In summary: Consider your sequence as a function.I am not allowed to. It has to be proven without any reference to functions.
  • #1
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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  • #2
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
 
  • #3
Hi,
Except that no one there was able to help me. Perhaps someone could now.
 
  • #4
Consider your sequence as a function.
 
  • #5
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.
 
  • #6
Write out what a_(n+1) actually is.
 
  • #7
sqrt(n+1) + 1/(n+1).
 
  • #8
Now what do you do to prove one sequence is larger than the other for all n≥2.
 
  • #9
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.
 
  • #18

1. How do you define a monotone sequence?

A monotone sequence is a sequence of numbers that either increases or decreases consistently without any fluctuations. This means that each term in the sequence is either greater than or equal to the previous term (for an increasing sequence) or less than or equal to the previous term (for a decreasing sequence).

2. What is the difference between a strictly monotone sequence and a non-strictly monotone sequence?

A strictly monotone sequence is a sequence in which each term is strictly greater than (or strictly less than, in the case of a decreasing sequence) the previous term. A non-strictly monotone sequence allows for the possibility of equal terms.

3. How do you prove that a sequence is monotone?

To prove that a sequence is monotone, you must show that it satisfies the definition of a monotone sequence. This can be done by analyzing the pattern of the sequence and showing that it consistently increases or decreases without any fluctuations. Alternatively, you can use mathematical induction to show that the sequence satisfies the definition for all values of n.

4. Why is it important to prove that a sequence is monotone?

Proving that a sequence is monotone is important because it allows us to make conclusions about the behavior of the sequence without having to examine each term individually. For example, if a sequence is proven to be increasing, we know that each term will be greater than or equal to the previous term, which can be helpful in making predictions or solving problems.

5. Can a sequence be both increasing and decreasing?

No, a sequence cannot be both increasing and decreasing. This would mean that the terms of the sequence are both greater than and less than the previous term, which is a contradiction. However, a sequence can be non-monotone, meaning it does not consistently increase or decrease, but rather has a mixture of increasing and decreasing terms.

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