On this page we'll give some hints for proving that the infinite series mentioned in the main article converge to the values claimed (using slightly more advanced maths).

We start with the series $$S=(2\times 1/2)+(3\times 1/4)+(4\times 1/8)+...+(n\times 1/2^{(n-1)})+...$$ First note that the series $$1/2+ 1/4+ 1/8+ ... + 1/2^n + ... $ converges to 1 (that's a standard result).

Using the ratio test you can prove that the series

$$S=(2\times 1/2)+(3\times 1/4)+(4\times 1/8)+...+(n\times 1/2^{(n-1)})+...$$

converges, in fact it converges absolutely. We're allowed to rearrange the terms of an absolutely convergent series, as this won't change its limit.

We'll rearrange the terms of $S$ in a way that splits it up into infinitely many individual series: $$\begin{array}{rclrclrclrclrc}S& =& 2/2& +& 3/4& +& 4/8& +& 5/16& ...& +& n/2^{n-1}& +& ...\\ & =& 1/2& +& 1/4& +& 1/8& +& 1/16& ...& +& 1/2^{n-1}& +& ...\\ & +& 1/2& +& 1/4& +& 1/8& +& 1/16& ...& +& 1/2^{n-1}& +& ...\\ & +& & & 1/4& +& 1/8& +& 1/16& ...& +& 1/2^{n-1}& +& ...\\ & +& & & & +& 1/8& +& 1/16& ...& +& 1/2^{n-1}& +& ...\\ & +& & & & +& & +& 1/16& ...& +& 1/2^{n-1}& +& ...\\ & +& & & & +& & +& & ...& +& ...& +& ...\end{array}$$ Each of these infinitely many series is of the form $$1/2^i+1/2^{i+1}+1/2^{i+2}+...$$ for some $i>1.$ It's not too hard to prove that such a sum converges to $1/2^{i-1}:$ $$1/2^i+1/2^{i+1}+1/2^{i+2}+...=1/2^{i-1}.$$ Substituting this into our expression for $S$ above we get $$\begin{array}{rclrclrclrclrc}S& =& 2/2& +& 3/4& +& 4/8& +& 5/16& ...& +& n/2^{n-1}& +& ...\\ & =& 1& & & & & & & & & & & \\ & +& 1& & & & & & & & & & & \\ & +& 1/2& & & & & & & & & & & \\ & +& 1/4& & & & & & & & & & & \\ & +& 1/8& & & & & & & & & & & \\ & +& ...& & & & & & & & & & & \end{array}$$ In other words, $$\begin{array}{rclr}S& =& 1+1+1/2+1/4+1/8+...+1/2^n+...& \\ & =& 1+1+1=3\end{array}$$ Now, in reality, women are unable to have infinitely many children. Let's assume that the maximal number of children a woman can have is $m.$ The probability of a woman having $n & m/2^{m-1} + (m+1)/2^{m} + (m+2)/2^{m+1}+ ... .\end{eqnarray*} This means that $$\begin{array}{rclrc}A& =& (1\times 0)+(2\times 1/2)+...+((m-1)\times 1/2^{m-2})& +& (m\times 1/2^{m-2})\\ & & (1\times 0)+(2\times 1/2)+...+((m-1)\times 1/2^{m-2})& +& (m\times 1/2^{m-1})+((m+1)\times 1/2^m)+((m+2)\times 1/2^{m+1}+...\\ & =& S=3.\end{array}$$ Thus, $3$ is an upper bound for the average number of children, even if we assume that the number of children a woman can have is at most $m.$ For the second box, notice that the series $s$ converges to $1$ as mentioned above. The fact that the infinite series converges to 1 can be by breaking it up into a sum of infinite series, as we did for $S.$ Now suppose that a woman can have at most $m$ children. The average number of sons is then $$A_s=1\times 1/2+1\times 1/4+1\times 1/8+...+1\times 1/2^{m-1}d=1.$$ The average number of daughters is $$A_d=(0\times 1/2)+(1\times 1/4)+(2\times 1/8)+...+((m-1)\times 1/2^m)+(m\times 1/2^m).$$ The last terms stands for women who have $m$ girls and then stop as they have reached the maximal number of children they can have. In a similar vein as above, notice that, by (1) above $$\begin{array}{rcl}m/2^m& =& m(1/2^{m+1}+1/2^{m+2}+...)\\ & & m/2^{m+1}+(m+1)/2^{m+2}+....\end{array}$$ This means that \begin{eqnarray*}A_d & = & (0 \times 1/2) + (1 \times 1/4) + (2 \times 1/8) + ... + ((m-1) \times 1/2^m) & +& (m \times 1/2^m) \ & & (0 \times 1/2) + (1 \times 1/4) + (2 \times 1/8) + ... + ((m-1) \times 1/2^m) & + & (m\times 1/2^{m+1}) + ((m+1)\times 1/2^{m+2})+.... \ & = & d = 1\end{eqnarray*) So both the average number of daughters and the average number of boys are less than $1,$ making the average number of children less than $2$.

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