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Exercise 1506Calculus › Single-variable calculus › Series › Power seriesL2Added 2 Nov 2026

The generating function of central binomial coefficients

#binomial series#generalised binomial coefficients#radius of convergence

Statement

1. Show that for ∣x∣<14|x|<\frac14,
11−4x=∑n≥0(2nn)xn.\frac1{\sqrt{1-4x}}=\sum_{n\ge0}\binom{2n}nx^n.
2. Deduce the radius of convergence of ∑(2nn)xn\sum\binom{2n}nx^n and the value of ∑n≥0(2nn)15n\sum_{n\ge0}\binom{2n}n\frac1{5^n}.

Hint

The binomial series is (1+u)α=∑n≥0(αn)un(1+u)^\alpha=\sum_{n\ge0}\binom\alpha nu^n with (αn)=α(α−1)⋯(α−n+1)n!\binom\alpha n=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}. Take α=−12\alpha=-\frac12 and u=−4xu=-4x.

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Solution

1. Coefficients. For α=−12\alpha=-\frac12,
(−1/2n)=(−12)(−32)⋯(−2n−12)n!=(−1)n(2n−1)!!2nn!,\binom{-1/2}n=\frac{\left(-\frac12\right)\left(-\frac32\right)\cdots\left(-\frac{2n-1}2\right)}{n!}=\frac{(-1)^n(2n-1)!!}{2^nn!},
where (2n−1)!!=1⋅3⋅5⋯(2n−1)=(2n)!2nn!(2n-1)!!=1\cdot3\cdot5\cdots(2n-1)=\frac{(2n)!}{2^nn!}. Hence
(−1/2n)=(−1)n(2n)!4n(n!)2=(−1)n4n(2nn).\binom{-1/2}n=\frac{(-1)^n(2n)!}{4^n(n!)^2}=\frac{(-1)^n}{4^n}\binom{2n}n.
With u=−4xu=-4x, (−1/2n)un=(2nn)xn\binom{-1/2}nu^n=\binom{2n}nx^n, which proves the formula for ∣4x∣<1|4x|<1.

2. Radius. The binomial series has radius 11 in uu, so R=14R=\frac14 in xx. This agrees with (2n+2n+1)(2nn)→4\frac{\binom{2n+2}{n+1}}{\binom{2n}n}\to4 and the root test result (2nn)1/n→4\binom{2n}n^{1/n}\to4.

Value. At x=15<14x=\frac15<\frac14:
∑n≥0(2nn)15n=11−45=5.\sum_{n\ge0}\binom{2n}n\frac1{5^n}=\frac1{\sqrt{1-\frac45}}=\sqrt5.