# Superpolynomial

This post is just to help a friend that wanted to see why $\displaystyle \sum_{i=1}^{\lfloor\sqrt{n}\rfloor} \binom{n}{i}$

was superpolynomial.

### Getting a simple lower bound

We can start by keeping only the last term. As all the terms are positive, that will give us a lower bound: $\displaystyle \sum_{i=1}^{\lfloor\sqrt{n}\rfloor} \binom{n}{i} \ge \binom{n}{\lfloor\sqrt{n}\rfloor}$.

We can expand the binomial coefficient in terms of factorials $\displaystyle \binom{n}{\lfloor\sqrt{n}\rfloor} = \frac{n!}{(n - \lfloor\sqrt{n}\rfloor)!\lfloor\sqrt{n}\rfloor!}$,

and some of the factors inside $n!$ can then be cancelled with the ones in $(n - \lfloor\sqrt{n}\rfloor)!$, leaving the following expression: $\displaystyle \binom{n}{\lfloor\sqrt{n}\rfloor} = \frac{n(n-1)\hdots(n - \lfloor\sqrt{n}\rfloor + 1)}{\lfloor\sqrt{n}\rfloor!}$.

As we want a lower bound, we can bound the numerator by below and the denominator by above, leaving us a bound without factorials: $\displaystyle \frac{n(n-1)\hdots(n - \lfloor\sqrt{n}\rfloor + 1)}{\lfloor\sqrt{n}\rfloor!} \ge \frac{(n - \sqrt{n})^{\sqrt{n} - 1}}{\sqrt{n}^{\sqrt{n}}}$.

For any reasonably big value of $n$ we will have $n - \lfloor\sqrt{n}\rfloor \ge n/2$. If we combine that with factoring out the -1 in the exponent we get $\displaystyle \frac{(n - \sqrt{n})^{\sqrt{n} - 1}}{\sqrt{n}^{\sqrt{n}}} \ge \frac{1}{n - \sqrt{n}}\left(\frac{n}{\sqrt{n}}\right)^{\sqrt{n}} \ge \frac{1}{n}\sqrt{n}^{\sqrt{n}}$,

removing the problematic floors and arriving to a much simpler expression.

Combining all these bounds we get $\displaystyle \sum_{i=1}^{\lfloor\sqrt{n}\rfloor} \binom{n}{i} \ge \frac{1}{n}\sqrt{n}^{\sqrt{n}}$,

a much simpler lower bound when compared to the original expression.

### Proving it is superpolynomial

If w take an arbitrary polynomial $p(n)$ of degree $k$ we can say for $n \ge 1$ that $\displaystyle p(n) = \sum_{i=1}^k p_i\,n^i \le \left(\sum_{i=1}^k |p_i|\right)n^k$,

meaning we can bound by above any polynomial of degree $k$ by a polynomial of the same degree with a single term.

Taking the limit of the quotient of logarithms, $\displaystyle \lim_{x \to +\infty} \frac{\log \frac{1}{n}\sqrt{n}^{\sqrt{n}}}{\log a\,n^k} = \lim_{x \to +\infty} \frac{\frac{\sqrt{n}}{2}\log n - \log n}{\log a + k\,\log n}$ $\displaystyle = \lim_{x \to +\infty} \frac{\log n\left(\frac{\sqrt{n}}{2} - 1\right)}{\log n\left(\frac{\log a}{\log n} + k\right)}$ $\displaystyle = \lim_{x \to +\infty} \frac{\frac{\sqrt{n}}{2} - 1}{\frac{\log a}{\log n} + k}$ $\displaystyle = +\infty$

we can see the lower bound of the original expression is superpolynomial, making the original expression superpolynomial too.