Sequence Limit: $\lim\limits_{n \rightarrow \infty}{n\,x^n}$

Question

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.

Answer 1

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.

Answer 2

Hint: if $|x| < r < 1$, show that $|(n+1) x^{n+1}| < r\, |n x^n|$ for sufficiently large $n$.

Answer 3

If $x=0$ we're done. Otherwise, we can show absolute convergence using l'Hôpital's, $$\begin{eqnarray*} \lim_{n\to\infty}|n x^n| &=& \lim_{n\to\infty} \frac{n}{|x|^{-n}} \\ &=& \lim_{n\to\infty} \frac{\frac{d}{dn} n}{\frac{d}{dn} |x|^{-n}} \\ &=& \lim_{n\to\infty} \frac{1}{-|x|^{-n}\log |x|} \\ &=& \lim_{n\to\infty} -\frac{|x|^n}{\log |x|} \\ &=& 0. \end{eqnarray*}$$

Answer 4

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 $r$ $$a_{n+1}=\frac{n+1}{r^{n+1}}=\frac{1}{r}\frac{n}{r^n}+\frac{1}{r}\frac{1}{r^n}=$$ $$=\frac{1}{r}a_n+\frac{1}{r}\frac 1 n a_n<\frac{1}{r}a_n+\frac{1}{r}a_n=\frac{2}{r}a_n$$

Answer 5

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.