√n + 1/n monotone proof

  • Thread starter peripatein
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  • #1
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Hi,

Any idea how I may prove that √n + 1/n is monotone? I am unable to show that √(n+1) + 1/(n+1) > √n + 1/n. Thanks!
 

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  • #2
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Hi,

Any idea how I may prove that √n + 1/n is monotone? I am unable to show that √(n+1) + 1/(n+1) > √n + 1/n. Thanks!
Are you required to use induction? That seems to be what you're doing.

A different approach would be to show that if f(x) = x1/2 + x-1, then f'(x) > 0 if x is large enough (it doesn't have to be very large).
 
  • #3
Ray Vickson
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Hi,

Any idea how I may prove that √n + 1/n is monotone? I am unable to show that √(n+1) + 1/(n+1) > √n + 1/n. Thanks!

The result is true if you mean √(n + (1/n)) and false if you mean what you *wrote*, which means (√n) + (1/n) (although in that case it is true for n ≥ 2).

RGV
 
  • #4
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Regardless of induction, an ascending monotone sequence satisfies the following:
There exist n0 natural, so that for every n>n0, the n+1th term of the sequence is greater than the nth term.
And I am not allowed to use functions.
 
  • #5
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Regardless of induction, an ascending monotone sequence satisfies the following:
There exist n0 natural, so that for every n>n0, the n+1th term of the sequence is greater than the nth term.
And I am not allowed to use functions.
Why not? You need to show us the complete problem statement.
 
  • #6
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The reason I am not allowed to use functions as we haven't dealt with them in class yet.
 
  • #7
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But surely you have studied functions in a previous class. The function f(x) = √x + 1/x is defined for real x > 0, and gives the same values at positive integers as the sequence sn = s(n) = √n + 1/n. Looking at the graph of f can give you some insight into what your sequence does.
 
  • #8
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As I am not permitted to use functions, could someone please suggest how I may prove that the above sequence is monotone?
 
  • #9
uart
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Regardless of induction, an ascending monotone sequence satisfies the following:
There exist n0 natural, so that for every n>n0, the n+1th term of the sequence is greater than the nth term.
And I am not allowed to use functions.

As Ray has already pointed out, your series is not monotonic. Just calculate the first three terms to demonstrate this.

Or were you given a more specific problem statement?
 
  • #10
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It is monotone, as for any n>=2 a_n+1>a_n
 
  • #11
uart
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It is monotone, as for any n>=2 a_n+1>a_n

Good, so the question I was asking is whether you were given "for n>=2" as part of the problem statement or if you were supposed to find for what range of "n" it was monotone. In any case, proceed as follows.

I am unable to show that √(n+1) + 1/(n+1) > √n + 1/n.
Start by rearranging the reciprocals on one side of the inequation and the square roots on the other. Simplify the reciprocals to a common denominator and then multiply each side by the conjugate surd, [itex](\sqrt{n+1} + \sqrt{n})[/itex]. It's relatively easy after that.
 

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