I answered a question Calculating entropy Through Huffman Codeword lengths. on this website and then got myself another question. If the alphabet has $5\,$ letters 'A', 'B', 'C', 'D', 'E' occuring with probabilities $\,0.4,\,0.2,\,0.2,\,0.1,\,0.1$, respectively, a letter-by-letter Huffman code gives an average code length of $\,L_1=2.2\,$ bits per letter. This number is greater than the Shannon information
per letter. My question is: if I group $n$ letters into one word and construct a word-by-word Huffman code, does the average code length per letter $\,L_n/n\rightarrow S\,$ approach Shannon's information content in the $\,n\rightarrow\infty\,$ limit? Is there a proof?
I wrote a C++ program to test this numerically. I assume the occurrence of letters are independent, i.e., the language is a random mixture of the $5$ letters with their individual probabilities. The result is plotted in the graph below:
The red dashed line is Shannon's information content $\,S\,$ per letter, which is unbeatable in this model as the letter occurrence is truely random. The blue curve is the average code length $\,L_n/n\,$ per letter of the Huffman tree for $\,n=1,2,\ldots,10$, where $L_n$ is the weighted tree height of the Huffman tree with $5^n$ leaf nodes. The numerical result does not clearly show whether $L_n/n\rightarrow S\;$ or not, as the curve is not monotonic. So a mathematical proof or disproof would be very much appreciated.