Happy New Year!... and polynomials

From remainder theorem, we note that $Q_n$ is the remainder of $P_n(2016)$ . Therefore,

$Q_n = P_n(2016) = (2016-1)(2016-2)...(2016-n) = \dfrac {2015!}{(2015-n)!}$

Therefore,

$\displaystyle \sum_{n=1}^{2015} {\dfrac {Q_n}{2015!}} = \sum_{i=1}^{2015} {\dfrac {1}{(2015-n)!}} = \sum_{m=0}^{2014} {\dfrac {1}{m!}}$

We know that:

$\displaystyle \sum_{m=0}^{2014} {\dfrac {1}{m!}} < \sum_{m=0}^{\infty} {\dfrac {1}{m!}} = e = 2.718281828 < \boxed {3}$

Set $y = x-2016$ . Rewrite $P_{n}(x)$ in terms of $y$ :

$\displaystyle P_{n}(x) = (y+2015)(y+2014)\ldots(y+2016-n) = \prod_{i=1}^{2016-n} (y+i)$

Multiplying out all the $i$ 's gives the constant term $\displaystyle Q_{n} = \prod_{i=1}^{2016-n} i = \frac{2015!}{(2016-n)!}$ .

From this, we know $\displaystyle S = \sum_{n=1}^{2015} \frac{Q_{n}}{2015!} = \sum_{n=1}^{2015} \frac{1}{(2016-n)!} = \sum_{n=1}^{2015} \frac{1}{n!}$ by noticing certain symmetries in the sum.

It is a well-known fact that $\displaystyle 3 > e = \sum_{n=1}^{\infty} \frac{1}{n!} > \sum_{n=1}^{2015} \frac{1}{n!}$ ; thus, by looking at the first few partial sums and noticing that the infinite series converges moderately quickly, we can conclude

$\lceil S \rceil = \boxed{3}$