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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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- Thread starter peripatein
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- #1

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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!

- #2

Mark44

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Are you required to use induction? That seems to be what you're doing.

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!

A different approach would be to show that if f(x) = x

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Ray Vickson

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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

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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

Mark44

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Why not? You need to show us the complete problem statement.

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.

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The reason I am not allowed to use functions as we haven't dealt with them in class yet.

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Mark44

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uart

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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?

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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.

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.I am unable to show that √(n+1) + 1/(n+1) > √n + 1/n.

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