fomula for expressing a fraction as 9269a series of factorial terms

Question

5/7= a series of terms with factorials as denominators

Answer 1

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)...x^r/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)...x^r/r! for r≥0 (0!=1). Every term in the expansion has a factorial in the denominator.

√2=(10/7)∑(-1)^r½(½-1)...x^r/r! (because 10/7=2(5/7)). So we have a power series for √2.