Putnam 1999 A4

Sum the series \[ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m^2n}{3^m(n3^m + m3^n)}. \]

That denominator looks pretty ugly and it doesn't seem like there's much of a way to simplify it right away. One thing to notice, though, is that it's nearly symmetric--if we interchange $m$ and $n$, we get something that looks pretty similar.

So what happens if, for any $x$ and $y$, we add up the terms where $(m,n) = (x,y)$ and $(m,n) = (y,x)$? (We should probably assume that $x\ne y$ for now; we'll have to handle those terms separately.) Let's see: that's \[ \frac{x^2y}{3^x(y3^x + x3^y)} + \frac{y^2x}{3^y(x3^y+y3^x)} = \frac{xy}{x3^y+y3^x} \left( \frac{x}{3^x} + \frac{y}{3^y} \right). \] That looks nice and symmetric.

Actually, it looks like we can get more things that look like $\frac{x}{3^x}$ if we divide the top and bottom of the left fraction by $3^x3^y$. That gets us \[ \frac{(x/3^x)(y/3^y)}{(x/3^x) + (y/3^y)} \left( \frac{x}{3^x} + \frac{y}{3^y} \right) = \frac{x}{3^x} \frac{y}{3^y}. \] Well, that's really simple.

Now we have to be careful. What do we actually get in the sum? If $x=y$ then we get $\frac{x^3}{2x3^{2x}} = \frac{x^2}{2 \cdot 3^{2x}}$, and for each pair of unequal $x$ and $y$ we get $\frac{x}{3^x} \frac{y}{3^y}$. So let's call the sum $S$; then we're saying \[ S = \sum_{x=1}^{\infty} \frac{x}{2 \cdot 3^x} + \sum_{x < y} \frac{x}{3^x} \frac{y}{3^y},\] and now we can play a slightly fancy trick that might not be obvious: if we want to change that second sum to sum over all pairs $(x,y)$ where $x \ne y$, we can notice that since the summand is symmetric, we'd just double our contribution. In other words, \[ \sum_{x \ne y} \frac{x}{3^x} \frac{y}{3^y} = 2 \sum_{x < y} \frac{x}{3^x} \frac{y}{3^y}. \] We're ready to put it all together now: \[ S = \sum_{x=1}^{\infty} \frac{x^2}{2 \cdot 3^x} + \sum_{x \ne y} \frac12 \frac{x}{3^x} \frac{y}{3^y} \], and we can bring the $x=y$ terms back into the fold, since $\frac12 \frac{x}{3^x} \frac{y}{3^y} =\frac{x^2}{2 \cdot 3^{2x}}$ when $x=y$. Finally: \[ S = \frac12 \sum_{x=1}^{\infty} \sum_{y=1}^{\infty} \frac{x}{3^x} \frac{y}{3^y} = \frac12 \left( \sum_{x=1}^{\infty} \frac{x}{3^x} \right) \left( \sum_{y=1}^{\infty} \frac{y}{3^y} \right),\] so the answer is just $T^2/2$, where $T = \sum_{n=1}^{\infty} \frac{n}{3^n}$.

If you're still with me, you probably know how to compute this sum. It's a calculus thing, really. In fact $\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}$ for $|x|<1$; start with the geometric series, take the derivative, and multiply by $x$. So plugging in $x=1/3$ gives $T = 3/4$ and hence $S = 9/32$.

It's not a bad idea to check a few terms to see if this looks like the right order of magnitude  for $S$; it's easy to lose a factor of $2$ here or there. This problem is not too easy, and it requires some attention to detail to make sure you're counting everything right. If you want to see what a very clean version of this same solution looks like, check the Putnam solutions page. It's usually a lot easier to write a very pretty and terse version of a solution once you've understood the original long-winded one you came up with, but we don't have time for refinements--there are five other problems to do and that one might have taken us an hour!

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