Proof involving a series?

We want to show that for |x| < 1,

(1 + x)^a = Σ(n = 0 to ∞) C(a, n) x^n,

where C(a, 0) = 1, and C(a, n) = a(a - 1)...(a - n+1)/n! for n > 0.

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Proving this with Lagrange Remainder formula is dreadful!.

Try this instead:

Let g(x) = Σ(n = 0 to ∞) C(a, n) x^n.

By the Ratio Test, this converges (at least) for |x| < 1:

r = lim(n→∞) |C(a, n+1) x^(n+1) / [C(a, n) x^n]|

..= |x| * lim(n→∞) |C(a, n+1) / C(a, n)|

..= |x| * lim(n→∞) |[a(a-1)...(a-n+1)(a-n) / (n+1)!] / [a(a-1)..(a-n+1)/n!]|

..= |x| * lim(n→∞) |(a-n) / (n+1)|

..= |x| * |-1|

..= |x|.

So, the series converges when r = |x| < 1.

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Next, we differentiate f in its region of convergence:

g'(x) = Σ(n = 1 to ∞) n C(a, n) x^(n-1)

.......= Σ(n = 0 to ∞) (n+1) C(a, n+1) x^n, by re-indexing

.......= Σ(n = 0 to ∞) (n+1) * [a(a-1)...(a-n+1)(a-n)/(n+1)!] x^n

.......= Σ(n = 0 to ∞) (a - n) * [a(a-1)...(a-n+1)/n!] x^n

.......= Σ(n = 0 to ∞) (a - n) * C(a, n) x^n

.......= a Σ(n = 0 to ∞) C(a, n) x^n - Σ(n = 0 to ∞) n C(a, n) x^n

.......= a g(x) - x g'(x).

Hence, (1 + x) g'(x) = a g(x)

==> g'(x)/g(x) = a/(1 + x).

Integrating both sides:

ln g(x) = a ln(1 + x) + A

==> g(x) = C(1 + x)^a, where C = e^A.

To find C, note that g(0) = 1 (from the definition of C).

So, g(0) = 1 = C * 1 ==> C = 1.

Hence, g(x) = (1 + x)^a, and we are done.

I hope this helps!