Number Theory

The first thing we do is move the a₇/7! to the other side:

$\frac57-\frac{a_7}{7!}=\frac{5\cdot6!}{7!}-\frac{a_7}{7!}=\frac{5\cdot6!-a_7}{7}\frac1{6!},$

we require the first fraction to be an integer. This is the case for a₇=2, now we can bring the next fraction to the left:

$\frac{514}{6!}-\frac{a_6}{6!}=\frac{514-a_6}{6}\frac{1}{5!},$

for the first fraction to be an integer a₆ needs to equal 4. We bring the next fraction to the left:

$\frac{85}{5!}-\frac{a_5}{5!}=\frac{85-a_5}{5}\frac1{4!},$

which requires a₅ to be zero. We bring the next fraction to the left:

$\frac{17}{4!}-\frac{a_4}{4!}=\frac{17-a_4}{4}\frac1{3!},$

which requires a₄ to be 1. We continue on our course:

$\frac{4}{3!}-\frac{a_3}{3!}=\frac{4-a_3}3\frac12,$

obviously a₃ should be one. We are left with:

$\frac12=\frac{a_2}{2!},$

and we get a₂=1, case solved. Checking our answer with the calculator:

$\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{4}{6!}+\frac{2}{7!}-\frac57=0.0000000000000000143,$

which is purely a machine error, since 1/7! equals 0.0001984126984127 which is far larger than our error.

Note that   5/7   =  0.7142857     [ the  7142857 is a repeating string ]

Try this  :

 a₂ = 1, a₃  = 1, a₄  = 1,  a₅  = 0, a₆ = 4, a₇ = 2

1/2! + 1/3! + 1/4! + 0/5! + 4/6! + 2/7!   =     5 / 7

I didn''t really have a mathematical way to solve this.....but, note  that

1/2!  + 1/3! + 1/4!  + 1/5! = 43/60 > 5/7

And 

1/2!  + 1/3!   + 1/4!  = 17/24  <  5/7

So......we need that    5/7  - 17/24   =  1/168   =   a₆/6! + a₇/7!

So........it was just a case of  letting a₅ = 0 and then  manipulating a₆  and a₇  to find a solution

Perhaps some other mathematician can find a more elegant way to the solution....!!!!