 Solution

For the maximum entropy distribution on the nonnegative integers, use the geometric distribution counting failures before the first success. Write $r=a/(a+1)$ and defineThe geometric series gives ${\sum }_i{g}_i=1$. Termwise differentiation of a power series applied to that geometric series for $0<r<1$ gives ${\sum }_{i}i{r}^i=r/(1-r{)}^2$, hence ${\sum }_{i}i{g}_i=r/(1-r)=a$.

For any competing probability mass function $p$ of mean $a$, its cross-entropy against $g$ isThis is finite. The fact that finite cross entropy bounds countable information entropy must be justified without assuming in advance that $H(p)$ is finite. Use $\ln u\le u-1$ with $u=g_i/p_i$ when $p_i>0$. Including zero terms by their limiting convention yields, for every finite $N$,Let $N\to \infty$. The information entropy partial sums increase, the cross-entropy partial sums converge to the finite value above, and the final mass difference tends to zero. This proves finiteness and the upper bound. We may now writeRelative entropy nonnegativity has equality exactly when $p=g$. Therefore the displayed geometric probability mass function is the unique maximizer, andThe mean constraint is essential: without it, uniform distributions on longer initial intervals have unbounded information entropy.