if $|a|<1$ so $\lim_{n\to \infty}na^n=0$. Prove that if $|a|<1$ ($a$ is real) so $\lim_{n\to \infty}na^n=0$.
I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.
thanks
 A: If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $\frac{n+1}{n}|a| < q < 1, \forall n \geq N$. Hence $n|a|^n = \frac{n}{n-1}|a|\frac{n-1}{n-2}|a|\dots\frac{N+1}{N}|a| \cdot N|a|^N < q^{n-N} \cdot N|a|^N, \, \forall n\geq N.$
The latter converges to zero and you get your squeezing sequence(s).
A: If $|a| < 1,$ we claim that the series $\sum_{1}^{\infty}na^{n}$ converges, so that
$na^{n} \to 0$. It suffices to show that the series is absolutely convergent. But, since
$$\frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+\frac{1}{n})|a| \to |a| < 1,$$ by the ratio test the series converges. Thus we have
$$\lim_{n \to \infty}na^{n} = 0.$$
A: You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,x\rightarrow\infty$ you can use L.Hospital theorem, $\lim \frac{x}{a^{-x}}=\lim \frac{1}{-a^{-x}}\frac{1}{lna}=0 $ 
A: Here a very elementary way:


*

*$|a| < 1 \stackrel{a \neq 0}{\Rightarrow} |a| = \frac{1}{1+p}$ for a $p > 0$.

*$(1+p)^n = 1 + np + \color{blue}{\frac{n(n-1)}{2}p^2} + \cdots + p^n \color{blue}{>\frac{n(n-1)}{2}p^2}$
So, it follows
$$0 \leq |na^n| = n|a|^n = \frac{n}{(1+p)^n} < \frac{n}{\color{blue}{\frac{n(n-1)}{2}p^2}} = \frac{2}{(n-1)p^2} \stackrel{n \to \infty}{\longrightarrow} 0$$
