A question on a proof technique of $\lim_{n\rightarrow \infty} n x^n = 0,\,\, x\in [0,1]$ So I try to prove the following statement $\lim_{n\rightarrow \infty} n x^n = 0, \quad x\in [0,1]$.
My question is if it is possible to prove this limit using d'alembert ratio test on the series $\sum_{n} nx^n \, \text{with } x \in [0,1]$ to see that the series is convergent and then simply applying the fact that for convergent series the coeficients in the limit tend to zero?
I feel like using d'alembert is a bit much here and as if there must be some bounding possible to squeeze the limit down to zero in the limit.
Thanks in advance!
 A: The ratio test certainly implies $nx^n\to 0$ when $0\leq x<1.$
If you think it is too big a gun to apply, you can make a similar argument.
Let $a_n=nx^n.$ Let $N>0$ be such that $\frac1N<1-x.$ This is equivalent to: $x\frac{N+1}{N}<1.$
Then for $n\geq N,$ $$a_{n+1} =\frac{n+1}n xa_n\leq\frac{N+1}Nx a_n\leq a_n$$
Then, for $n \geq N,$ $a_n$ is decreasing and non-negative, so it must have a limit $L.$
But then taking the limit of both sides of $a_{n+1}=\frac{n+1}nx a_n$ gives $L=xL,$ and since $x\neq 1,$ $L=0.$

This is just repeating a core part of the ratio test, but without any knowledge about the series convergence. Basically, if $|x|<\lim_{n\to\infty}\left|\frac{b_{n+1}}{b_n}\right|$, then $|b_nx^n|$ is bounded above by some geometric sequence $Cy^n$ with $0\leq y<1.$
If $\left|\frac{b_{n}}{b_{n+1}}\right|\to R$ and $|x|<R,$ pick $\epsilon=R-|x|.$ Then there exists and $N$ such that $$\left|R-\left|\frac{b_{n}}{b_{n+1}}\right|\right|<\epsilon$$ for all $n\geq N.$
But that means $\left|\frac{b_{n}}{b_{n+1}}\right|>|x|$ for all $n\geq N.$
From there, we similarly show that $a_n=|b_nx^n|$ is decreasing for $n\geq N,$ and take the limit of both sides of both sides of $a_{n+1}=\left|\frac{b_{n+1}}{b_{n}}\right|xa_n$ to get $L=\frac{|x|}{R}L,$ and $|x|/R<1$ means $L=0.$

You can do likewise with the stronger root test.
If $b_n$ has $\limsup |x|\sqrt[n]{|b_n|} =K<1,$  we can show there is a $C$ such that, for large $n,$ $$|b_nx^n|\leq CK^n\to 0.$$
A: Here's a direct way, just using the binomial expansion. Let $y=\frac{1}{x}-1$ Then $y>0$ and $x=\frac{1}{y+1}$. $$nx^n=\frac{n}{(y+1)^n}$$ $$<\frac{6n}{y^3n(n-1)(n-2)}$$ $$<\frac{24}{y^3n^2}\text { for } n>4$$ $$<\frac{24}{n} \text { if }y^3n>1$$Thus, for $ \epsilon>0, $ if $n>\max(4,\frac{1}{y^3},\frac{24}{\epsilon})$, then $nx^n<\epsilon.$
