TITLE: Determining bounds for the sum $\sum\limits_{n=1}^\infty \frac{1}{2^n - 3^n }$
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: The sum $$S = \sum_{n\ge 1} \frac{1}{3^n-2^n}$$ is a special evaluation of a more general harmonic sum and may be treated by Mellin transforms. The base function $f(x)$ is $$f(x) = \frac{1}{3^x-2^x}$$ which has the following Mellin transform:
$$\mathfrak{M}(f(x); s) = \int_0^\infty \frac{1}{3^x-2^x} x^{s-1} dx =
\int_0^\infty \frac{1}{3^x} \frac{1}{1-(2/3)^x} x^{s-1} dx \\=
\int_0^\infty \frac{1}{3^x} \sum_{q\ge 0} \left(\frac{2}{3}\right)^{qx} x^{s-1} dx =
\sum_{q\ge 0} \int_0^\infty \left(\frac{2^q}{3^{q+1}}\right)^x x^{s-1} dx \\ =
\sum_{q\ge 0} \int_0^\infty e^{(q\log(2/3)-\log 3)x} x^{s-1} dx =
\Gamma(s) \sum_{q\ge 0} \frac{1}{(\log 3 - q\log(2/3))^s} \\=
\Gamma(s) \sum_{q\ge 0} \frac{1}{(\log 3 + q\log(3/2))^s} =
\frac{\Gamma(s)}{(\log(3/2))^s}  \sum_{q\ge 0} \frac{1}{(\log (3)/\log (3/2) + q)^s} \\ =
\frac{\Gamma(s)}{(\log(3/2))^s}  \zeta\left(s,\log (3)/\log (3/2) \right). $$
It follows that the Mellin transform $g^*(s)$ of the harmonic sum
$$ g(x) = \sum_{n\ge 1} \frac{1}{3^{nx}-2^{nx}}$$ is given by
$$ g^*(s) = \mathfrak{M}(g(x); s) = 
\frac{\Gamma(s)}{(\log(3/2))^s}  \zeta\left(s,\log (3)/\log (3/2) \right) \zeta(s).$$
Looking at the first few poles we perform Mellin inversion to get
$$ \operatorname{Res}(g^*(s)/x^s; s = 1) =
-\frac{\log x + \psi(\log (3)/\log (3/2)) + \log\log(3/2)}{x\log(3/2)} \\
\operatorname{Res}(g^*(s)/x^s; s = 0) =
\frac{1}{4} \frac{2\log(3)-\log(3/2)}{\log(3/2)} \\
\operatorname{Res}(g^*(s)/x^s; s = -1) =
-{\frac {1}{144}}\,{\frac {x \left(  \left( \log  \left( 3/2 \right)  \right) ^{2}
+6\, \left( \log  \left( 3 \right)  \right) ^{2}-6\,\ln  \left( 3 \right) \log
 \left( 3/2 \right)  \right) }{\log  \left( 3/2 \right) }} \\
\operatorname{Res}(g^*(s)/x^s; s = -3) \\=
{\frac {1}{86400}}\,{\frac {{x}^{3} \left( - \left( \log  \left( 3/2 \right)
 \right) ^{4}+30\, \left( \log  \left( 3 \right)  \right) ^{2} \left( \log  \left(
3/2 \right)  \right) ^{2}+30\, \left( \log  \left( 3 \right)  \right) ^{4}-60\,
 \left( \log  \left( 3 \right)  \right) ^{3}\log  \left( 3/2 \right)  \right) }{
\log  \left( 3/2 \right) }}.$$
Finally set $x=1$ and switch to numerics, including a few more terms, to obtain the approximation
$$ S \approx 0.2508048329 + 1.104755646  - 0.08106948472 + 0.0004958137737 \\
- 0.5355881093 \times 10^{-5} +  0.6368945126 \times 10^{-7} = 1.274981516.$$
This agrees with the exact value to the extend of the precision used.