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I am looking for a clarification on this answer about why $\lim_{n\to\infty}nx^n=0$.

Specifically, the answer has two aspects I am concerned about: First, the answer begins by saying

We prove the result under the slightly weaker condition $|x|\lt 1$.

Let $|x|=\dfrac{1}{1+t}$. Then $t\gt 0$.

Why can we let $\vert x\vert = \frac{1}{1+t}$?,

  • And $t>0$ because $\vert x\vert$ must be positive, correct?

Second, the answer claims

By the Binomial Theorem, if $n \ge 2$, then $$(1+t)^n \ge 1+nt +\frac{n(n-1)}{2}t^2 \gt \frac{n(n-1)}{2}t^2.$$

Why is the first inequality weak? We are leaving off terms from the binomial formula, and $t>0$, so shouldn't it be $>$ (strict)?

Note: The original answerer has not be active in over a year, which is why I post here rather than leave a comment.

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user106860
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    In that first case you can actually just solve for $t$, but even if you couldn't do this algebraically you can still make such a substitution for a monotone function by using the intermediate value theorem. As for the second thing, since $t>0$ the inequality is strict as soon as $n>2$. This is really nitpicking though because the strictness isn't needed for the argument at all. – Ian May 25 '18 at 23:41
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    "Why can we let |x|=1/1+t?". Because if $0< |x| < 1$ then $\frac 1{|x|} > 1$. And $\frac 1{|x|} - 1 > 0$. So let $t = \frac 1{|x|} - 1; t > 0$ then $|x| = \frac 1{1+t}; t> 0$. – fleablood May 25 '18 at 23:42
  • "We are leaving off terms from the binomial formula, and t>0, so shouldn't it be > (strict)?" Only if $n > 2$ if $n = 2$ it's an equality. – fleablood May 25 '18 at 23:44
  • "And t>0 because |x| must be positive, correct?" Not really. $t > 0$ because we are assuming $|x| < 1$. – fleablood May 25 '18 at 23:45
  • @Ian Sorry, you lost me when talking about substituting for monotone function. You are just saying that since $\frac{1}{1+t}$ is monotone, I can find values of $t$ such that $\frac{1}{1+t} <\vert x\vert$ and another value $t'\not=t$ such that $\frac{1}{1+t}>\vert x\vert$ and then by IVT there is a $t''$ such that $\frac{1}{1+t} =\vert x\vert$? – user106860 May 26 '18 at 00:18
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    @user106860 Yes, that's right, and $t''$ is unique as a solution to the equation. – Ian May 26 '18 at 02:19
  • @Ian Thank you c – user106860 May 26 '18 at 02:23

1 Answers1

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  1. $t$ has to be non-negative because $\;0<|x|=\dfrac1{1+t}<1$, so the denominator has to be $>1$.
  2. The first inequality is weak because it's an equality if these are the only terms of the binomial expansion, which happens if $n=2$.
Bernard
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