perfect power

The power $m^n$ is called a perfect power if $m$ and $n$ are integers both greater than 1. The perfect powers form the ascending order sequence (cf. Sloane’s http://oeis.org/classic/A001597A001597)

$4,\mathrm{\hspace{0.17em}8},\mathrm{\hspace{0.17em}9},\mathrm{\hspace{0.17em}16},\mathrm{\hspace{0.17em}25},\mathrm{\hspace{0.17em}27},\mathrm{\hspace{0.17em}32},\mathrm{\hspace{0.17em}36},\mathrm{\hspace{0.17em}49},\mathrm{\hspace{0.17em}64},\mathrm{\hspace{0.17em}81},\mathrm{\hspace{0.17em}100},\mathrm{\hspace{0.17em}121},\mathrm{\hspace{0.17em}125},\mathrm{\dots },$(1)

i.e.

$$2^2,{\mathrm{\hspace{0.33em}2}}^3,{\mathrm{\hspace{0.17em}3}}^2,{\mathrm{\hspace{0.33em}2}}^4=4^2,{\mathrm{\hspace{0.33em}5}}^2,{\mathrm{\hspace{0.33em}3}}^3,{\mathrm{\hspace{0.33em}2}}^5,{\mathrm{\hspace{0.17em}6}}^2,{\mathrm{\hspace{0.33em}7}}^2,{\mathrm{\hspace{0.33em}2}}^6=4^3=8^2,{\mathrm{\hspace{0.33em}3}}^4=9^2,{\mathrm{\hspace{0.33em}10}}^2,{\mathrm{\hspace{0.33em}11}}^2,{\mathrm{\hspace{0.33em}5}}^3,\mathrm{\dots }$$

S. S. Pillai has conjectured in 1945, that if the $i^{\mathrm{th}}$ member of the sequence (1) is denoted by $a_i$, then

$\underset{i\to \mathrm{\infty }}{lim\; inf}(a_{i+1}-a_i)=\mathrm{\infty }.$(2)

This does not necessarily that one had  ${lim}_{i\to \mathrm{\infty }}(a_{i+1}-a_i)=\mathrm{\infty }$,  since there may always exist little differences $a_{i+1}-a_i$ arbitrarily far from the begin of the sequence (1).

The equation (2) is equivalent (http://planetmath.org/Equivalent3) to the

Pillai’s conjecture.  For any positive integer $k$, the Diophantine equation

$$x^m-y^n=k$$

has only a finite number of solutions  $(x,y,m,n)$  where the integers $x,y,m,n$ all are greater than 1.

Pillai’s conjecture generalises the Catalan’s conjecture ($k=1$) in which the number of solutions is 1.

The series formed by the inverse numbers of the perfect powers converges absolutely, and its sum may be calculated easily:

${\displaystyle \sum _{m,n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^n}}$$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \sum _{n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^n}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \sum _{n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}\cdot {\displaystyle \frac{1}{m^{n-2}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}{\displaystyle \sum _{n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^{n-2}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left({\displaystyle \frac{1}{m}}\right)}^n$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}\cdot {\displaystyle \frac{1}{1-\frac{1}{m}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m(m-1)}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{m-1}}-{\displaystyle \frac{1}{m}}\right)$

The sum of this telescoping series (http://planetmath.org/TelescopingSum) is equal to 1.