Determining bounds for the sum $\sum\limits_{n=1}^\infty \frac{1}{2^n - 3^n }$

Question

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.

Answer 1

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 $1$.

Answer 2

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 $$ \sum\limits_{n\geqslant1}\frac1{x_n}\geqslant1+\frac15=\frac65. $$

Answer 3

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.

Answer 4

$\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.