# √n + 1/n monotone proof

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!

Mark44
Mentor
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).

Ray Vickson
Homework Helper
Dearly Missed
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

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.

Mark44
Mentor
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.

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

Mark44
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.

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

uart
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?

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

uart
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.