Odds in Pascal's Triangle Let $O(n)$ be the number of odds in rows $0-n$ in Pascal's triangle. Let $E(n)$ be the number of evens in rows $0-n$. I have heard the claim that the $\lim_{n \to \infty} \frac{O(n)}{E(n)}=0$. Does anyone have a proof of this? It seems like the relationship between Pascal's Triangle and the binary representation of the row number could be useful, but I am not seeing it. 
 A: I think it's easier to show the equivalent claim that $\lim \frac{O}{O+E}=0$.
Let $b(n)$ be the number of $1$ in the binary expansion of $n$. Then it is a well-known fact (alluded to in the question) that there are $2^{b(n)}$ odd numbers in the $n^{\rm th}$ row of Pascal's triangle, and so $O(n)=\sum_{i=0}^n 2^{b(n)}$.
We will start by computing $O(2^n-1)$. There are $\binom{n}{k}$ numbers less than $2^n$ that have $k$ $1$s in their binary expansion. Summing over all $k$, we have
$O(2^n-1)=\sum_{i=0}^{2^{n-1}} 2^{b(n)}=\sum_{k=0}^n 2^k \binom{n}{k}=3^n$.
On the other hand, $O(2^{n-1}-1)+E(2^{n-1}-1)$ is the total number of entries in the  first $2^{n-1}-1$ rows, and thus is equal to $\frac{2^{n-1}(2^{n-1}+1)}{2}=2^{n-2}(2^{n-1}+1)$. This grows asymptotically like $4^n$, so clearly $\lim_{n\to \infty} \frac{O(2^n-1)}{O(2^{n-1}-1)+E(2^{n-1}-1)}=0$.
But $O$ and $O+E$ are increasing functions. Thus, for any $l$, we may take $n$ with $2^{n-1} \leq l \leq 2^n - 1$, and then $\frac{O(l)}{O(l)+E(l)} < \frac{O(2^{n}-1)}{O(2^{n-1}-1)+E(2^{n-1}-1)}$. As the latter sequence converges to $0$ and the former sequence is positive, it too must converge to $0$.
A: If you look at the structure of the odd elements in Pascal's triangle you'll see a well-known self-similar figure essentially equivalent to the Sierpinsky Triangle.  Once you prove the self-similarity, you can use it to bound the ratio $R(2^n) = \frac{O(2^n)}{E(2^n)}$ by $\left(\frac34+o(1)\right)R(2^{n-1})$ (and in particular, bound it by e.g. $\frac45R(2^{n-1})$ for sufficiently large $n$); this provides a subsequence that converges to $0$.  The other part of the proof is to show that the 'intermediate' ratios (i.e., $R(k)$ for $2^{n-1}\lt k \lt 2^n$) never get too much larger; this is a little more complicated, but still relatively straightforward.
