February, 2017

Solution:

Let $Q(x)=x^{2} P(x)-1$. Then $Q(n)=n^{2} P(n)-1=0$ for $n=1,2, \ldots, 2016$, and $Q$ has degree $2017$. Thus we may write
$$Q(x)=x^{2} P(x)-1=(x-1)(x-2) \ldots(x-2016) L(x)$$
where $L(x)$ is some linear polynomial. Then $Q(0)=-1=(-1)(-2) \ldots(-2016) L(0)$, so $L(0)=-\frac{1}{2016!}$.

Now note that
$$\begin{aligned} Q^{\prime}(x) & =x^{2} P^{\prime}(x)+2 x P(x) \\ & =\sum_{i=1}^{2016}(x-1) \ldots(x-(i-1))(x-(i+1)) \ldots(x-2016) L(x)+(x-1)(x-2) \ldots(x-2016) L^{\prime}(x) \end{aligned}$$
Thus
$$Q^{\prime}(0)=0=L(0)\left(\frac{2016!}{-1}+\frac{2016!}{-2}+\ldots+\frac{2016!}{-2016}\right)+2016!L^{\prime}(0)$$
whence $L^{\prime}(0)=L(0)\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2016}\right)=-\frac{H_{2016}}{2016!}$, where $H_{n}$ denotes the $n$th harmonic number.

As a result, we have $L(x)=-\frac{H_{2016} x+1}{2016!}$. Then
$$Q(2017)=2017^{2} P(2017)-1=2016!\left(-\frac{2017 H_{2016}+1}{2016!}\right)$$
which is $-2017 H_{2016}-1$. Thus
$$P(2017)=\frac{-H_{2016}}{2017}$$
From which we get $2017 P(2017)=-H_{2016}$. It remains to approximate $H_{2016}$. We alter the well known approximation
$$H_{n} \approx \int_{1}^{n} \frac{1}{x} d x=\log x$$
to
$$H_{n} \approx 1+\frac{1}{2}+\int_{3}^{n} \frac{1}{x} d x=1+\frac{1}{2}+\log (2016)-\log (3) \approx \log (2016)+\frac{1}{2}$$
so that it suffices to lower bound $\log (2016)$. Note that $e^{3} \approx 20$, which is close enough for our purposes. Then $e^{6} \approx 400 \Longrightarrow e^{7} \approx 1080$, and $e^{3} \approx 20<2^{5} \Longrightarrow e^{0.6} \ll 2 \Longrightarrow e^{7.6}<2016$, so that $\log (2016)>7.6$. It follows that $H_{2016} \approx \log (2016)+0.5=7.6+0.5>8$ (of course these are loose estimates, but more than good enough for our purposes). Thus $-9<2017 P(2017)<-8$, making our answer $-9$.