Prove that $f_{n}:[1,+\infty[ \rightarrow \mathbb{R} : x \rightarrow \frac{x^n}{1+x^{n}}e^{-x}$ is increasing By definition, I want to prove that $f_{n+1}(x)-f_{n}(x)\ge0$ for $x \in [1,+\infty[$

So, we obtain : $e^{-x}(\frac{x^{n+1}-x^{n}}{(1+x^{n+1})(1+x^{n})})$

But for $x \in [1,+\infty[$ we have : $e^{-x}&gt;0$, and also $(1+x^{n+1})(1+x^{n})&gt;0$ for all $n \in \mathbb{N}.$

and we know that : the sequence $(x^n)_{n \in \mathbb{N}}$ is increasing which means that :

$x^{n+1}-x^{n}\ge 0$ so $f_n$ is increasing.

Is this method correct ? 

Thanks in advance.
No, you are on the wrong trace. To prove that $f_n(x)$ is increasing, you should pick $x_1&lt;x_2$ and show that $f(x_1)\le f(x_2)$.

Or you can verify that $f_n'(x)&gt;0$.
No, you are on the wrong trace. To prove that $f_n(x)$ is increasing, you should pick $x_1&lt;x_2$ and show that $f(x_1)\le f(x_2)$.

Or you can verify that $f_n'(x)&gt;0$.