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Wolfram alpha gives that this is $0$, but I'm not sure how to show it.

I Tried writing as $\frac{(1-x)^n}{1/n}$ and using L'Hopital's rule, but a new $n$ term shows up every time I take the derivative.

I also try to set $h=\frac1n$ and then write as $$ \lim_{n\to\infty} n(1-x)^n = \lim_{h\to 0} \frac{(1-x)^\frac1h}{h} $$ but same problem, numerator goes to $0$, denominator to $0$, and L'Hôpital doesn't seem to work because the derivative of the numerator gives $\frac1h (1-x)^{\frac1h -1}$, which has the initial trouble

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    $\begingroup$ ratio test?${}$ $\endgroup$ – Lord Shark the Unknown May 26 '18 at 21:27
  • $\begingroup$ @LordSharktheUnknown Ratio converges to $<1$ so series converges so the sequence of terms must go to $0$? $\endgroup$ – user106860 May 26 '18 at 21:51
  • $\begingroup$ "Lord Shark the Unknown" has posted an astute comment, and several people have posted answers solving the problem by being "technical". Being "technical" is necessary on many occasions, but actually understanding is the ultimate goal (and those for whom it is not the goal should do something worthwhile instead). I posted an answer explaining not just how to prove that the limit is $0,$ but why that is what you should expect. $\endgroup$ – Michael Hardy May 26 '18 at 21:53
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Incrementing $n$ by $1$ has the effect of multiplying $n$ by $\dfrac{n+1} n,$ making it bigger, but also has the effect of multiplying $(1-x)^n$ by $(1-x),$ making it smaller. Which will prevail?

As $n$ grows, it ultimately gets immensely bigger than $\dfrac{1-x} x.$

When that happens, incrementing $n$ by $1$ has the effect of multiplying $n(1-x)^n$ by $\left( 1 + \dfrac 1 n \right)(1-x),$ which is a number less than $1$ and which gets smaller as $n$ grows.

So after reaching some large value of $n$, the rest of the sequence is bounded above by a geometric sequence with a common ratio less than $1;$ thus is approaches $0.$

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  • $\begingroup$ Is the geometric sequence just $\sum (1-x)^n$? (or I maybe a constant times that? since it is only after a large enough value of $n$). I guess I am wondering if you can state that last sentence "so after reaching..." more formally, if it is not too much work. $\endgroup$ – user106860 May 26 '18 at 21:53
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    $\begingroup$ @user106860 : The ratio is something bigger than $1-x,$ but if you go to a larger value of $n,$ i.e. further down the sequence, it gets closer to $1-x.$ You can make it as close as you want to $1-x$ by going far enough along the sequence, since you can make $1 + \dfrac 1 n$ as close to $1$ as you want by making $n$ big enough. $\qquad$ $\endgroup$ – Michael Hardy May 26 '18 at 21:57
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Take the series $\;\sum\limits_{n=1}^\infty n (1-x)^n\;$ and apply the $\;n\,-$ th root test to its absolute value:

$$\sqrt[n]{n|1-x|^n}=\sqrt[n]n\,|1-x|\xrightarrow[n\to\infty]{}|1-x|$$

and we thus get absolute convergence iff

$$|1-x|<1\implies -1<1-x<1\implies -2<-x<0\implies 0<x<2$$

so the series' sequence's general term tends to zero.

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We have

$$n(1-x)^n=e^{\log n+n\log (1-x)}\to 0$$

indeed

$$\log n+n\log (1-x)=n\left(\frac{\log n}n+\log (1-x)\right)\to -\infty$$

since $\frac{\log n}n\to 0$ and $\log (1-x)<0$.

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You can solve this by taking the logarithm of each side. Let $y = \lim_{n \rightarrow \infty} n(1-x)^n$, and then we see that $$\ln(y) = \ln(\lim_{n \rightarrow \infty} n(1 -x)^n) = \lim_{n\rightarrow \infty } \ln(n (1-x)^n)$$ by using the continuity of the natural log function. Now, we use log rules and see $$\ln(y) = \lim_{n \rightarrow \infty} \ln(n) + n\ln(1-x).$$ Since $1 - x$ is less than 1, $\ln(1-x) < 0$. The $\ln(n)$ term grows logarithmically in $n$ and the $n\ln(1-x)$ term grows linearly in $n$. So $\ln(y)$ tends to $- \infty$, hence $y = e^{\ln(y)}$ tends to 0.

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    $\begingroup$ Please: Write $\ln(y)$ or $\ln y,$ not $ln(y). \qquad$ $\endgroup$ – Michael Hardy May 26 '18 at 21:37
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I mean, if you really want to use L'Hôpital, you can write it as $$ \frac{n}{(1-x)^{-n}}, $$ and then differentiating numerator and denominator gives $$ \frac{1}{-n(1-x)^{-n-1}} = \frac{-1}{1-x} \frac{(1-x)^n}{n}, $$ and the right hand side tends to $0$ since both variable terms in the product do.

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    $\begingroup$ Minor error in derivative of denominator. Exponent should be -(n+1). $\endgroup$ – herb steinberg May 26 '18 at 21:36

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