Generating Function Sum and Combinotorics

Question

Problem: The sum $\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$ has a finite value. Determine that value.

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I am quite stuck on how to do this. Can somebody give me <only> a hint or hints to get going? Thanks!

Answer 1

Hint: start repeatedly differentiating the generating function for $\frac{1}{1 - x}$.

Answer:

We have the generating function/Maclaurin series for $\frac{1}{1 - x}$: $$\frac{1}{1 - x} = 1 + x + x^2 + \ldots \quad |x| < 1$$ Differentiating, which we do term by term does not change the radius of convergence, giving us $$\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \ldots \quad |x| < 1.$$ Differentiating a second time gives us $$\frac{2}{(1 - x)^3} = (2 \cdot 1) + (3 \cdot 2)x + (4 \cdot 3)x^2 + \ldots \quad |x| < 1$$ Recall that $\binom{n}{2} = \frac{1}{2}n(n - 1)$, so we have $$\frac{1}{(1 - x)^3} = \binom{2}{2} + \binom{3}{2}x + \binom{4}{2}x^2 + \ldots \quad |x| < 1.$$ Dividing by $16$, and substituting $x = \frac{1}{4}$, we get $$\frac{4}{27} = \frac{1}{16} \cdot \frac{1}{\left(1 - \frac{1}{4}\right)^3} = \frac{\binom{2}{2}}{4^2} + \frac{\binom{3}{2}}{4^3} + \frac{\binom{4}{2}}{4^4} + \ldots$$

Answer 2

Consider $$S=\sum_{n=2}^\infty \binom{n}{2}x^n =\frac 12 \sum_{n=2}^\infty n(n+1) x^n=\frac 12 \sum_{n=2}^\infty[n(n-1)+2n] x^n$$ $$S=\frac 12 \sum_{n=2}^\infty n(n-1) x^n+ \sum_{n=2}^\infty n x^n=\frac {x^2}2 \sum_{n=2}^\infty n(n-1) x^{n-2}+x\sum_{n=2}^\infty n x^{n-1}$$ $$S=\frac {x^2}2 \left(\sum_{n=2}^\infty x^n \right)''+x\left(\sum_{n=2}^\infty x^n \right)'$$ When finished, let $x=\frac 14$.