<s>[INST] If $a_k = c^k-1$, where $c > 1$, what can be said about $s(n) =\sum_{k=1}^n \frac{a_k}{a_{k+1}} $? 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}
$
 [/INST] 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 $s(n)\geq \frac{n}{c}-\frac{1}{c(c+1)}-\frac{1}{(c-1)(c^2+c+1)}$.
</s>