<s>[INST] Determining bounds for the sum $\sum\limits_{n=1}^\infty \frac{1}{2^n - 3^n }$ I have to give low and high bounds for the following:
$$
\sum_{n=1}^\infty \frac{1}{2^n - 3^n }
$$
How do I determine an upper bound? How can I show this sum exists?
edit: removed erroneous conclusion.
 [/INST] $\begin{align}
\sum_{n=1}^\infty \frac{1}{3^n - 2^n }
&=\sum_{n=1}^\infty \frac{1}{3^n(1 - (2/3)^n) }\\
&=\sum_{n=1}^\infty \frac{1}{3^n} \sum_{k=0}^\infty(2/3)^{nk}) \\
&=\sum_{n=1}^\infty \frac{1}{3^n}+\sum_{n=1}^\infty \frac{1}{3^n} \sum_{k=1}^\infty(2/3)^{nk}) \\
&= \frac{1/3}{1-1/3}
+ \sum_{m=1}^{\infty} (2/3)^m \sum_{d|m} (1/3)^d\\
&= \frac{1}{2}
+ \sum_{m=1}^{\infty} (2/3)^m \sum_{d|m} (1/3)^d\\
\end{align}
$
Looking at the inner sum,
let
$S(m, r) = \sum_{d|m}r^d$,
where $0 < r < 1$.
$S(m, r) > r$,
since the term $d=1$ always occurs.
$S(m, r) < \sum_{d=1}^{m}r^d
= \frac{r}{1-r}
$,
since more terms are here than
in the original sum.
So
$S(m, 1/3) > 1/3$
and
$S(m, 1/3) < 1/2$.
Therefore
$\begin{align}
\sum_{n=1}^\infty \frac{1}{3^n - 2^n }
&= \frac{1}{2}
+ \sum_{m=1}^{\infty} (2/3)^m S(m, 1/3)\\
&> \frac{1}{2}
+ \sum_{m=1}^{\infty} (2/3)^m (1/3)\\
&> \frac{1}{2}
+ \frac1{3}\sum_{m=1}^{\infty} (2/3)^m\\
&= \frac{1}{2}
+ \frac1{3}\frac{2/3}{1-2/3}\\
&= \frac{1}{2}
+ \frac1{3}2\\
&= \frac{1}{2}
+ \frac{2}{3}\\
&= \frac{7}{6}\\
\end{align}
$
and
$\begin{align}
\sum_{n=1}^\infty \frac{1}{3^n - 2^n }
&< \frac{1}{2}
+ \frac1{2}2\\
&= \frac{3}{2}
\end{align}
$
More accurate bounds for
$S(m, r)$
would result in more accurate bounds
for the sum.
</s>