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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.

J. M. ain't a mathematician
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jon jones
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6 Answers6

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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.

lhf
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  • But the series test is just the same for sequences. In fact, if $\lim;\left| \dfrac{a_{n+1}}{a_n}\right|<1$ then $a_n \to 0$. – Pedro Jul 20 '12 at 01:42
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    @Ihf, I like your answer but in order to be able to use it I think you must put $,n|x|^n,$, as the ratio series is for positive series. Anyway, +1 – DonAntonio Jul 20 '12 at 02:50
  • @DonAntonio, but if a series converges absolutely, then it converges. See Wikipedia for this version of the ratio test. – lhf Jul 20 '12 at 10:59
  • Oh, I know that, @Ihf, but the ratio test is appliable only to positive series, that's all. My point is that, since $,-1<x<1,$, the test cannot be applied to the series as it is. Of course, once you take $,|nx^n|,$ and prove convergence, you have absolute conv. and thus conv. – DonAntonio Jul 20 '12 at 14:00
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Hint: if $|x| < r < 1$, show that $|(n+1) x^{n+1}| < r\, |n x^n|$ for sufficiently large $n$.

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

user26872
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    I don't think this is strictly correct for $x<0$. You differentiate with respect to $n$, treating it as a continuous variable. But for $x<0$, the quantity $x^n$ is not necessarily real for $n \not \in \mathbb{N}$. – James Fennell Jul 20 '12 at 09:00
  • @JamesFennell: Thanks, James. I believe you are right. – user26872 Jul 20 '12 at 14:55
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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 $k>0$.

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

Pedro
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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.

N. Mao
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