Show that $\forall q \in (1,2)$, $\exists n \in N^+$ let there is no $m \in N^+$ satisfy the inequality$$q^n-q^{n-1}+1 \le q^m\lt q^{n+1}-q^n+1$$
The inequality can be written as $$q^{n-1}\le\frac{q^m-1}{q-1}\lt q^n$$
The sum of the geometric sequence is bigger than the $n-1$th term and the $n$th
term of a geometric sequence with a common ratio of $q$.
But I don't know what I can do next.
I am not sure whether the statement is correct.
On
Let $\epsilon_n \in [0,1] \forall n$ with $\epsilon_{t+1}>\epsilon_t$.
We have that $q$ is expressible as $1+\epsilon_1$. This leads to: $$1+\epsilon_2-\epsilon_3\leq q^m < 1+\epsilon_3 - \epsilon_2$$ This draws that the only possible integer solution is $0$, which isn't positive.
On
$q^n-q^{n-1}+1 \le q^m \lt q^{n+1}-q^n+1 $
The right inequality implies $m \le n$ because $q^{n+1}-q^n+1 \lt q^{n+1} $ so $m < n+1$ so $m \le n$.
The left inequality, in contrast, has many solutions.
Write it as $q^{n-1}(q-1)+1 \le q^m $ and let $q = 1+x$ where $0 < x < 1$. It becomes $x(1+x)^{n-1}+1 \le (1+x)^m $.
If $\frac1{n} \le x \le \frac1{n-1}$, then $(1+x)^{n-1} \le (1+\frac1{n-1})^{n-1} \lt e $ so the left side is less than $1+\frac{e}{n-1} $.
For the right side, $q^m =(1+x)^m \ge 1+mx \ge 1+\frac{m}{n} $ so the left inequality is true if $1+\frac{e}{n-1} \lt 1+\frac{m}{n} $ or $m > \frac{en}{n-1} \gt 5 $ if $n \ge 2 $.
Therefore, for any $m > 5$ and $n > 2$ we can find a $q$ so that the left inequality holds.
Consider fixed arbitrary $q \in (1,2)$. Consider $q^i(2-q)$ as $i$ varies. This diverges to infinity, so in particular is eventually $\geq 1$. Also $q^0(2-q)=2-q<1$. Thus there exists a smallest integer $k$ with $q^k(2-q) \geq 1$, and we have $k \geq 1$. By minimality of $k$ we get $q^{k-1}(2-q)<1$.
Take $n=k \in \mathbb{N}^{+}$. We get $$q^{k-1}=q^{k-1}(q-1)+q^{k-1}(2-q)<q^{k-1}(q-1)+1=q^k-q^{k-1}+1$$ $$q^k=q^k(q-1)+q^k(2-q) \geq q^k(q-1)+1=q^{k+1}-q^k+1$$ We are thus done, as $q^k-q^{k-1}+1 \leq q^m<q^{k+1}-q^k+1$ then implies $k-1<m<k$, a contradiction, so no such $m$ exists.
Remark: In fact it is possible to show that $q(2-q)<1$ for $q\in (1,2)$, thus $k \geq 2$, which reconciles with the fact that $m=1$ always works when $n=1$.