 Solution

Put $S={\sum }_{k=1}^a{2}^{-l_k}$ and let $l_{\max }$ be the largest length. Expanding $S^n$ sums $2^{-L}$ over all strings of $n$ source symbols, where $L$ is the total code length. By unique decodability, the concatenations of length $r$ in this sum are all distinct, so there are at most $2^r$ of them. Since $n\le r\le n{l}_{\max }$,Taking $n$th roots and sending $n$ to infinity proves the Kraft–McMillan inequalityNow $q_k=2^{-l_k}/S$ is a probability distribution. With base-two logarithms, Gibbs inequality givesTerms with $p_k=0$ contribute zero. ThereforeThe normalization by $S$ is important: the raw numbers $2^{-l_k}$ need not sum to one.