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.
 A: I strongly suggest not working with the sum as it stands. The terms are obviously negative. Change signs. Your intuition will get much better. For sure  mine would.
Find upper and lower bounds for the sign-altered sum, and then use the result to draw conclusions about upper and lower bounds for the original sum.
To show that the sign-altered sum exists, we can note that $2^n\le \frac{2}{3}3^n$, and therefore $\frac{1}{3^n-2^n} \le \frac{3}{3^n}$. From this you can also get an upper bound for $\sum_1^\infty \frac{1}{3^n-2^n}$. 
As for a lower bound for $\sum_1^\infty \frac{1}{3^n-2^n}$, a very easy one is $0$! Almost as easy is to use the first term, which is $1$.
A: Let $x_n=3^n-2^n$, then $2^n\leqslant2\cdot3^{n-1}=3^n-3^{n-1}$ hence $x_n\geqslant3^{n-1}$ for every $n\geqslant1$. Thus, every $x_n$ is positive and 
$$
\sum\limits_{n\geqslant1}\frac1{x_n}\leqslant\sum\limits_{n\geqslant0}\frac1{3^n}=\frac32.
$$
In particular, the series $\displaystyle\sum\limits_{n}\frac1{x_n}$ converges (absolutely) and its sum is at most $\dfrac32$. 
To get a lower bound, use $x_1=1$ and $x_2=5$ to get 
$$
\sum\limits_{n\geqslant1}\frac1{x_n}\geqslant1+\frac15=\frac65.
$$
A: $\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.
