2000 Putnam A1

Problem: Let $A$ be a positive real number. What are the possible values of $$\sum_{j=0}^{\infty} x_j^2,$$ given that $x_0,x_1,\dots$ are positive numbers for which $$\sum_{j=0}^{\infty}=A?$$

Solution: The answer is $\boxed{(0,A^2)}$. \\ There are two things we need to prove: first, that the possible values of $\sum_{j=0}^{\infty} x_j^2$ must lie in the interval $(0,A^2)$, and second, all values in this interval can be achieved. \\ For our first problem, the values are obviously positive. Also, note that $$\sum_{j=0}^{\infty} \frac{x_j}{A}=1,$$ and because $x_j>0$, we must have $$0 <  \frac{x_j}A   < 1,$$ which implies that $$\left(\frac{x_j}A\right)^2   <  \frac{x_j}A  ,$$$$\implies \sum_{j=0}^{\infty}\left(\frac{x_j}A\right)^2   <  \sum_{j=0}^{\infty}\left(\frac{x_j}A\right)   = 1.$$Multiplying by $A^2$, we get $$\sum_{j=0}^{\infty}x_j^2 < A^2.$$ \\ Now, for the second part, let the sequence of $x_j$'s be a geometric sequence with ratio $r$. Then, we have $$\sum_{j=0}^{\infty}x_j=\frac{x_0}{1-r},$$ so \begin{align*} &\sum_{j=0}^{\infty}x_j^2\\ &=\frac{x_0^2}{1-r^2}\\ &=\left (\frac{1-r}{1+r}\right )\left (\sum_{j=0}^{\infty} x_j \right )^2.\\ &=\left (\frac{1-r}{1+r}\right ) A^2. \end{align*}Now, notice that we can make $\frac{1-r}{1+r}$ take on any value in the range $(0,1)$ by taking the appropriate value of $r$ in $(0,1)$. In particular, $x=\frac{1-r}{1+r}$ implies $r=\frac{1-x}{1+x}$. As $r$ goes from $0$ to $1$, the fraction goes from $1$ to $0$. This proves that that all values in the interval $(0,A^2)$ can be achieved. \\ We have proved both sub-problems, and we are done. $\square$