 Solution

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.