5/7=5×6!/(7×6!)=3600/5040.
For AP (arithmetic progression) we will have a finite number n in the series:
a, a+d, a+2d, a+3d, ..., a+(n-1)d.
Sum of the series, S=(n/2)(a+a+(n-1)d)=(n/2)(2a+nd-d)=3600.
n(2a+nd-d)=7200,
n2d+n(2a-d)=7200,
n2+n(2a-d)/d=7200/d; complete the square:
n2+n(2a-d)/d+(2a-d)2/(4d2)=7200/d+(2a-d)2/(4d2),
(n+(2a-d)/(2d))2=7200/d+(2a-d)2/(4d2),
n+(2a-d)/(2d)=√(7200/d+(2a-d)2/(4d2)).
If we choose d=2a then this simplifies:
n=√(7200/d). If we choose d=8 (for example) n=√900=30.
Since d=8, a=4, and the series is: 4, 12, 20, ..., 228, 236.
The sum of the series is 3600 so to get the fraction 5/7 we need to divide by 7!=5040.
The final series is (1/7!){4,12,20,...,220,228,236} and its sum is 3600/5040=5/7.
The series can be written (1/7!)∑(4+8n) for 0≤n≤29, so each term contains 7! in its denominator.
5/7=4/7!+12/7!+20/7!+...+220/7!+228/7!+236/7!
There are many solutions to this problem, and we could have used an infinite geometric progression (GP) instead of an AP, for example:
5/7=(7/10)∑50-n for n≥0, which is an infinite series. However, this doesn't contain factorials in the denominators.
It's possible that this question has been misstated, because 5/7 is an approximation to 1/√2, and it can be used to generate an infinite series which does contain factorials in the denominators of the terms.
(5/7)2=25/49; ½=25/50=25/49-1/98=(25/49)(1-1/50).
√½=1/√2=√2/2=√[(25/49)(1-1/50)]=(5/7)(1-0.02)½.
(1-x)n can be expanded binomially:
1-nx+n(n-1)x2/2!-n(n-1)(n-2)x3/3!+...+∑(-1)rn(n-1)...xr/r!
Let n=½ and x=0.02 and we get:
(1-0.02)½=√0.98=√(98/100)=√[(49/100)×2]=(7/10)√2.
So (5/7)(1-0.02)½=(5/7)((7/10)√2)=√2/2=1/√2.
Therefore 1/√2=(5/7)∑(-1)r½(½-1)...xr/r! for r≥0 (0!=1). Every term in the expansion has a factorial in the denominator.
√2=(10/7)∑(-1)r½(½-1)...xr/r! (because 10/7=2(5/7)). So we have a power series for √2.