Sequence Limit: $\lim\limits_{n \rightarrow \infty}{n\,x^n}$ If $-1<x<1$ show that $\lim\limits_{n \to \infty}{n\,x^n} = 0$. I don't have idea. I only prove that $n\,x^n$ is decreasing. 
 A: Consider the series $\sum n x^n$. The ratio test shows that this series converges when $|x|<1$. Hence in this case the sequence of terms must converge to zero.
A: Hint: if $|x| < r < 1$, show that $|(n+1) x^{n+1}| < r\, |n x^n|$ for sufficiently large $n$.
A: If $x=0$ we're done. 
Otherwise, we can show absolute convergence using l'Hôpital's, 
$$\begin{eqnarray*}
\lim_{n\to\infty}|n x^n|
&=& \lim_{n\to\infty} \frac{n}{|x|^{-n}} \\
&=& \lim_{n\to\infty} \frac{\frac{d}{dn} n}{\frac{d}{dn} |x|^{-n}} \\
&=& \lim_{n\to\infty} \frac{1}{-|x|^{-n}\log |x|} \\
&=& \lim_{n\to\infty} -\frac{|x|^n}{\log |x|} \\
&=& 0.
\end{eqnarray*}$$
A: It's not restrictive to assume $x\ne0$, so $0<|x|<1$. Set $1/\sqrt{|x|}=1+y$, so $y>0$; by Bernoulli’s inequality,
$$
(1+y)^n>1+ny
$$
so
$$
|x|^n<\frac{1}{((1+y)^n)^2}<\frac{1}{(1+ny)^2}
$$
Therefore
$$
|nx^n|<\frac{n}{(1+ny)^2}
$$
Note that a similar technique proves that, when $|x|<1$,
$$
\lim_{n\to\infty}n^kx^n=0
$$
for every $k>0$.
A: If $-1<x<1$ then $x=\dfrac{1}{r}$, with $|r|>1$. For example:
$$1>0.1=\frac{1}{10}$$
$$1>0.25=\frac{1}{4}$$
$$1>0.\overline 3 =\frac{1}{3}$$
Then, we can write your sequence as
$$a_n=\frac{n}{r^n}$$
Can you try and see what would happen to $a_n$ for large $n$?
Say $r=2$. Then what would $$\lim\limits_{n\to \infty}\frac{n}{2^n}$$ be? Can you try and generalize?
Also note that for positive $r$ $$a_{n+1}=\frac{n+1}{r^{n+1}}=\frac{1}{r}\frac{n}{r^n}+\frac{1}{r}\frac{1}{r^n}=$$
$$=\frac{1}{r}a_n+\frac{1}{r}\frac 1 n a_n<\frac{1}{r}a_n+\frac{1}{r}a_n=\frac{2}{r}a_n$$
A: I believe that one could use the monotonicity of the functions of $n$ in order to justify taking the continuous derivative with respect to $n$ in the l'Hopital argument above (by oen). This would still only work for x nonnegative (by what James mentioned). But then, couldn't one argue that, because
$|nx^n| = |ny^n|$ for positive integers $n$ and for constants $x$ and $y$ (where $x=-y$ and $y$ is positive), that $\lim_{n\to\infty}|nx^n| = \lim_{n\to\infty}|ny^n| = \lim_{n\to\infty}ny^n = 0$. (The second to last equality is valid because $ny^n >0$ and the last equality is the limit whose value is given by the l'Hopital argument.) Could one then say that since $\lim_{n\to\infty}|nx^n| = 0$ that $\lim_{n\to\infty}nx^n = 0$? (-1 < x < 0) This seems fine to me, but let me know what you think. 
