How to solve $n\left(\frac{n}{n+1}\right)^n>1$ for $n\in\Bbb N$?

I tried to solve the inequality

$$n\left(\frac{n}{n+1}\right)^n>1,\quad n\in\Bbb N$$

with this code

Solve[x*(x/(1 + x))^x > 1 && x > 0, x, Integers]


but the program said

This system cannot be solved with the methods available to Solve.

My attempt using 'NSolve' failed too. My question: there is a way to solve the inequality?

Reduce works like a charm:

Reduce[x*(x/(1 + x))^x > 1 && x > 0, x, Integers]


x ∈ Integers && x >= 3

• Yes, this is exactly what I want, thank you. – Masacroso Jan 25 '17 at 20:42
• When I tried this myself I forgot the $x > 0$ constraint, and so MM returned an error. Now I'm kicking myself. Nicely done. – Michael Seifert Jan 25 '17 at 20:48
• Actually I was a little bit dissappointed that i had to add the constraint. I'd thought that mathematica itself would plug me the solution out with the additional constraint since the function is anyway complex between [-1,0] and everything below -1 is negative. But thanks. – Julien Kluge Jan 25 '17 at 20:59

One approach is to use FindRoot for the equality:

FindRoot[n (n/(n + 1))^n == 1, {n, 2.5}]

{n -> 2.29317}


So any n greater than 2.29 will have the correct direction in the inequality, i.e., n=3 would be the smallest integer.

If you only need to know a limited number of examples that solve the inequality, then FindInstance will work:

FindInstance[x (x/(1 + x))^x > 1, x, Integers, 2]
(* {{x -> 278}, {x -> 358}} *)


The 2 in the above code tells Mathematica to find two instances. The instances that Mathematica finds are pretty unpredictable (if you only ask it for one instance, it returns x -> 16), and it takes longer than one might want to run; but depending on the precise circumstances, it could be an option.