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For $|x|<1$, the binomial series is

For a complex number $\alpha$, set $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{\alpha }{n}\right)=\alpha (\alpha -1)\cdots (\alpha -n+1)/n!$ for $n\ge 1$. The displayed power series represents $\exp (\alpha \mathrm{Log}(1+z))$ on the unit disc, using the branch of the complex logarithm with value zero at $1$. It terminates for a nonnegative integer $\alpha$. Otherwise the ratio test gives absolute convergence for $|z|<1$. Termwise differentiation gives $(1+z)F^{\prime }(z)=\alpha F(z)$ and $F(0)=1$, which determine that holomorphic function. With $\alpha =-1/2$ and argument $-z$, its coefficients are the normalized central binomial coefficients.

The central binomial coefficientcounts the $n$-element subsets of a $2n$-element set. It is the middle and largest coefficient in the binomial expansion of $(1+x{)}^{2n}$, by unimodality of binomial coefficients. For $n=2$, the row $1,4,6,4,1$ has central value six.

Equivalently, $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ counts binary strings of length $2n$ with exactly $n$ ones. Dividing by the total $2^{2n}$ gives the probability of exactly $n$ successes in $2n$ independent fair Bernoulli trials. The normalized central binomial coefficients therefore connect a simple counting sequence to balanced random outcomes.

Extend the sequence by $c_0=1$. Cancelling factorials in consecutive normalized central binomial coefficients givesThe given logarithmic inequality therefore impliesOn $[j,j+1]$, $1/t\le 1/j$, so summing its integrals gives ${\sum }_{j=1}^{n}1/j\ge \log (n+1)$. Exponentiating yields the explicit boundAlso every consecutive ratio is less than one, so $(c_n)$ is strictly decreasing, a fact needed for the alternating endpoint.

Use the generalized binomial theorem with exponent $-1/2$, choosing the analytic branch of a square root that equals $1$ at $z=0$. For $|z|<1$,The signs cancel in each binomial coefficient. For $n\ge 1$,The same expression equals $1$ for $n=0$, using the empty-product convention. Consequently the power series isIts coefficients are the normalized central binomial coefficients. As a convergence check, the ratio of successive absolute terms tends to $|z|$, so the ratio test gives absolute convergence inside the unit disc. Applying the Cauchy coefficient formula to the holomorphic function $(1-z{)}^{-1/2}$ now immediately yieldsThis value is independent of the chosen radius because the contour integral extracts the same coefficient on every such circle.