How to prove $\lim_{n\rightarrow\infty}nx^n=0$ without L'Hôpital's rule, where $x \in [0,1)$??

Question

How to prove $$\lim_{n\rightarrow\infty}nx^n=0$$ without L'Hôpital's rule? where $x \in [0,1)$ and $n=1,2,3,...$.

I know one of way to prove this is to treat $n$ is real, and $n$ and $(\frac{1}{x})^n$ as functions of $n$. Then, apply L'Hôpital's rule and it works for discrete $n$ by the definiation of limit.

However, I would like to skip the step we treat $n$ is real. Is there any proof for this question only using the techniques for limit of sequences? (such as the sandwich theorem, etc.)

Answer 1

For $0 < x < 1$ we have $x = (1+y)^{-1}$ with $y > 0$.

It follows from the binomial theorem that,

$$ (1+y)^n > \frac{n(n-1)y^2}{2}.$$

Hence,

$$nx^n = \frac{n}{(1+y)^n}< \frac{2}{(n-1)y^2}.$$

For any $\epsilon > 0$, if $n > 1 + 2/(\epsilon y^2)$, then $|nx^n-0|= nx^n < \epsilon$.

Thus, $\lim_{n \to \infty} nx^n = 0$.

Answer 2

Hint: When $|x| < 1$ we have that $$\sum_{n=1}^{\infty} n x^n = \frac{x}{(x-1)^2}$$

Answer 3

Let $x \in ]0, 1[$ (the case where $x = 0$ is trivial). Since $$ \frac{(n+1)x^{n+1}}{nx^{n}} = (1+\frac{1}{n})x \to x < 1, $$ the series $\sum_{n} nx^{n}$ converges, so $nx^{n} \to 0$.

I hope this is helpful.

Answer 4

Let $x<1-1/N$. If $n>2N$ then $(n+1)x^{n+1}/(nx^n)<(1+\frac1{2N})(1-\frac1N)<1-\frac1{2N}$. So terms from then on shrink by at least that factor, and $nx^n<C(1-\frac1{2N})^n\to0$.

Answer 5

If $x\lt1$, then find $n_x$ so that $\frac1{n_x}\le\frac{1-x}2$. Then, for any $k\ge n_x$, $\frac1k\le\frac{1-x}2\implies x\le\frac{k-2}k$ and $$ \begin{align} \frac{(k+1)x^{k+1}}{kx^k} &=\frac{k+1}kx\\ &\le\frac{k+1}k\frac{k-2}k\\ &=1-\frac1k-\frac2{k^2}\\ &\le\frac{k-1}k \end{align} $$ Thus, for any $n\gt n_x$, $$ \begin{align} \frac{nx^n}{n_xx^{n_x}} &=\prod_{k=n_x}^{n-1}\frac{(k+1)x^{k+1}}{kx^k}\\ &\le\prod_{k=n_x}^{n-1}\frac{k-1}k\\ &=\frac{n_x-1}{n-1} \end{align} $$ Therefore, $$ \begin{align} \lim_{n\to\infty}nx^n &\le\lim_{n\to\infty}\frac{n_xx^{n_x}(n_x-1)}{n-1}\\[6pt] &=0 \end{align} $$ where the numerator is a fixed constant depending on $x\lt1$.

Answer 6

Let $0\lt x\lt 1$. Then there is a positive $a$ such that $x=\frac{1}{1+a}$. By the Binomial Theorem, we have for $n\ge 2$ that $$0\lt (1+t)^n =1+nt+\frac{n(n-1)}{2}t^2\gt \frac{n(n-1)}{2}t^2.$$ It follows that $$0\lt nx^{n} \lt \frac{2}{t^2}\cdot \frac{1}{n-1}.$$ Now it is straightforward, given a positive $\epsilon$, to produce an $N$ such that $0\lt nx^n \lt \epsilon$ if $n\gt N$.