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.