Clarify an old answer: Why we can let $\vert x \vert = \frac{1}{1+t}$ for $\vert x \vert <1$

Question

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.

Answer 1

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 $n=2$.