If $a_k = c^k-1$, where $c > 1$, what can be said about $s(n) =\sum_{k=1}^n \dfrac{a_k}{a_{k+1}} $?
This is a generalization of $a_n=3^n-1$, prove that $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots+\frac{a_n}{a_{n+1}}>\frac{n}{3}-\frac{1}{8}.$ which is the case $c=3$.
I can show that $\dfrac{n}{c}-\dfrac{1}{c^2-1} \lt s(n) \lt \dfrac{n}{c}-\dfrac1{c^2} $ and I wondered if better inequalities can be found.
Here is my method.
$\begin{array}\\ s(n) &=\sum_{k=1}^n \dfrac{a_k}{a_{k+1}}\\ &=\sum_{k=1}^n \dfrac{c^k-1}{c^{k+1}-1}\\ &=\sum_{k=1}^n \dfrac{c^k-1/c+1/c-1}{c^{k+1}-1}\\ &=\sum_{k=1}^n (\dfrac1{c}-\dfrac{1-1/c}{c^{k+1}-1})\\ &=\dfrac{n}{c}-\sum_{k=1}^n \dfrac{1-1/c}{c^{k+1}-1}\\ &=\dfrac{n}{c}-(1-\frac1{c})\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\ &=\dfrac{n}{c}-\dfrac{c-1}{c}r(n) \qquad\text{where }r(n)=\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\ r(n) &=\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\ &=\dfrac1{c^2-1}+\sum_{k=2}^n \dfrac{1}{c^{k+1}-1}\\ &>\dfrac1{c^2-1}-\dfrac1{c^2}+\dfrac1{c^2}+\sum_{k=2}^n \dfrac{1}{c^{k+1}}\\ &=\dfrac1{c^2(c^2-1)}+\sum_{k=1}^n \dfrac{1}{c^{k+1}}\\ &=\dfrac1{c^2(c^2-1)}+\dfrac{1-1/c^{n}}{c^2(1-1/c)}\\ &=\dfrac1{c^2(c^2-1)}+\dfrac{1}{c(c-1)}-\dfrac{1}{c^{n+1}(c-1)}\\ &=\dfrac{1}{c(c-1)}+\dfrac1{c^2(c^2-1)}(1-\dfrac1{c^{n-1}})\\ &\ge\dfrac{1}{c(c-1)}\\ \text{so}\\ s(n) &<\dfrac{n}{c}-\dfrac1{c^2}\\ \text{and}\\ r(n) &=\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\ &<\sum_{k=1}^n \dfrac{1}{c^{k+1}-c^{k-1}}\\ &<\sum_{k=1}^n \dfrac{1}{c^{k-1}(c^2-1)}\\ &=\dfrac{1}{c^2-1}\sum_{k=0}^{n-1} \dfrac{1}{c^{k}}\\ &<\dfrac{1}{c^2-1}\sum_{k=0}^{\infty} \dfrac{1}{c^{k}}\\ &=\dfrac{1}{(c^2-1)(1-1/c)}\\ &=\dfrac{c}{(c^2-1)(c-1)}\\ \text{so}\\ s(n) &>\dfrac{n}{c}-\dfrac{c-1}{c}\dfrac{c}{(c^2-1)(c-1)}\\ &>\dfrac{n}{c}-\dfrac{1}{c^2-1}\\ \end{array} $
Given the identity $$ s(n)=\frac{n}{c}-\frac{c-1}{c}\left[\frac{1}{c^2-1}+\sum_{k=2}^{n}\frac{1}{c^{k+1}-1}\right] $$ we have $s(n)\leq \frac{n}{c}-\frac{1}{c(c+1)}$ for any $n\geq 2$. Under the same assumption $$ \sum_{k=2}^{n}\frac{1}{c^{k+1}-1}\leq \sum_{k=2}^{+\infty}\frac{1}{c^{k+1}-1}\leq \sum_{k=2}^{+\infty}\frac{1}{c^{k+1}-c^{k-2}}=\frac{1}{c-\frac{1}{c^2}}\cdot\frac{1}{c(c-1)} $$ leads to $s(n)\geq \frac{n}{c}-\frac{1}{c(c+1)}-\frac{1}{(c-1)(c^2+c+1)}$.