Question

Prove that!


`\tiny\displaystyle\rm \sum_(q=1)^(\infty)\left(a^(q)\left(1+\sum_(r=1)^(\infty) (\prod \limits_(i=0)^(r+1)(q-i))/(\Gamma(r+1))\left((b x^(m))/(a)\right)^(r)\right)\right)=(a+b x^(m))/(1-\left(a+b x^(m)\right))`

if all variables belong to natural number.

Answer 1

Recall that if |x| < 1, then

`\displaystyle \frac1{1-x} = \sum_(k=0)^\infty x^k`

So if |a + bxᵐ| < 1, then

`\displaystyle (a+bx^m)/(1 - (a+bx^m)) = \sum_(q=1)^\infty (a+bx^m)^q = \sum_(q=1)^\infty a^q \left(1 + \frac{bx^m}a\right)^q`

By the binomial theorem,

`\displaystyle \left(1 + \frac{bx^m}a\right)^q = \sum_(r=0)^q \binom qr \left(\frac{bx^m}a\right)^r`

The binomial coefficient is defined as

`\dbinom nk = (n!)/(k!(n-k)!)`

if 0 ≤ k ≤ n, and 0 otherwise. Then we can rewrite the sum as

`\displaystyle \left(1 + \frac{bx^m}a\right)^q = \sum_(r=0)^\infty \binom qr \left(\frac{bx^m}a\right)^r`

and pulling out the first term,

`\displaystyle \left(1 + \frac{bx^m}a\right)^q = 1 + \sum_(r=1)^\infty \binom qr \left(\frac{bx^m}a\right)^r`

Finally,

`\dbinom qr = (q!)/(r! (q-r)!) \\\\\\ = (q(q-1)(q-2)\cdots(q-(r-1))(q-r)(q-(r+1))\cdots3\cdot2\cdot1)/(r! (q-r) (q-(r+1)) (q-(r+2))\cdots3\cdot2\cdot1) \\\\\\ = (q(q-1)(q-2)\cdots(q-(r-1)))/(r!) \\\\ = (\prod\limits_(i=0)^(r-1)(q-i))/(\Gamma(r+1))`

since n! = Γ(n + 1). (I think the given upper limit in the product may be a mistake.)