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If $-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.

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    Your acceptation rate converges to zero faster than any other sequence I've ever met...perhaps you'd want to fix this?2012-07-20

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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.

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    Oh, I *know* that, @Ihf, but the ratio test is appliable *only* to positive series, that's all. My point is that, since \,-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.2012-07-20
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Hint: if $|x| < r < 1$, show that $|(n+1) x^{n+1}| < r\, |n x^n|$ for sufficiently large $n$.

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

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    @JamesFennell: Thanks, James. I believe you are right.2012-07-20
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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 $k>0$.

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

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    @DonAntonio That should have read |r|>1. Fixing...2012-07-20
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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.