= Solution
Extend the <sequence> by $c_0=1$. Cancelling factorials in consecutive <normalized central binomial coefficients> gives
$$
\frac{c_n}{c_{n-1}}=\frac{(2n)(2n-1)}{4n^2}
=1-\frac1{2n},\qquad
c_n=\prod_{j=1}^n\left(1-\frac1{2j}\right).
$$
The given logarithmic inequality therefore implies
$$
\log c_n\leq-\frac12\sum_{j=1}^n\frac1j.
$$
On $[j,j+1]$, $1/t\leq1/j$, so summing its integrals gives $\sum_{j=1}^n1/j\geq\log(n+1)$. Exponentiating yields the explicit bound
$$
\boxed{0<c_n\leq(n+1)^{-1/2}\longrightarrow0.}
$$
Also every consecutive ratio is less than one, so $(c_n)$ is strictly decreasing, a fact needed for the alternating endpoint.
Back to article page
