# √n + 1/n monotone proof

1. Nov 2, 2012

### peripatein

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!

2. Nov 2, 2012

### Staff: Mentor

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. Nov 2, 2012

### Ray Vickson

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. Nov 2, 2012

### peripatein

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. Nov 3, 2012

### Staff: Mentor

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

6. Nov 3, 2012

### peripatein

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

7. Nov 3, 2012

### Staff: Mentor

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. Nov 5, 2012

### peripatein

As I am not permitted to use functions, could someone please suggest how I may prove that the above sequence is monotone?

9. Nov 5, 2012

### uart

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. Nov 5, 2012

### peripatein

It is monotone, as for any n>=2 a_n+1>a_n

11. Nov 5, 2012

### uart

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, $(\sqrt{n+1} + \sqrt{n})$. It's relatively easy after that.