Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What methods do you know for finding the limit of a sequence? –  Eric Stucky Jul 20 '12 at 0:51
    
more basic possible. –  jon jones Jul 20 '12 at 1:12
1  
Are you able to show that if $|r|<1$, then $r^n \to 0$? –  copper.hat Jul 20 '12 at 1:17
2  
It's not true that $n x^n$ is decreasing, although it is true that $|n x^n|$ is decreasing for large enough $n$. –  Robert Israel Jul 20 '12 at 1:27
6  
Your acceptation rate converges to zero faster than any other sequence I've ever met...perhaps you'd want to fix this? –  DonAntonio Jul 20 '12 at 2:48

5 Answers 5

up vote 4 down vote accepted

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.

share|improve this answer
    
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 Tamaroff Jul 20 '12 at 1:42
1  
@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 2:50
    
@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

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.

share|improve this answer

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

share|improve this answer
1  
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 9:00
    
@JamesFennell: Thanks, James. I believe you are right. –  user26872 Jul 20 '12 at 14:55

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

share|improve this answer

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

share|improve this answer
    
Where you wrote, in your first line, "...for $\,r>1\,$" , it must be, I believe, "...for $\,r>1\,\,\,or\,\,\,r<-1\,$ –  DonAntonio Jul 20 '12 at 2:52
1  
@DonAntonio That should have read $|r|>1$. Fixing... –  Pedro Tamaroff Jul 20 '12 at 2:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.