Is it right to say that $\left|\frac{x^n}{n+1}\right|\leq |x^n|=|x|^n$ if $n$ is a natural number $\geq 0$?
This is not an isolated question but it is related to this my Solution verification of $\sum_{k=1}^{\infty}\frac{x^n}{n+1}$ with $x\in\mathbb{R}$ previous question. I have tried to ask for this in comments without success so I hope this separated question could help me. Thanks in advance!
Yes, this is valid: because $f(x) = \frac1{x}$ is strictly decreasing, $n + 1\geq 1$ implies $\frac1{n+1} \leq 1,$ and because $|x| \geq 0$ for any $x$ we can say that $\left|\frac{x^n}{n+1}\right| = \left|\frac{1}{n+1}\right|\cdot|x|^n \leq |x|^n.$