300 Followers Problem - Polynomial Differential Reciprocal Summations!

Instead of $2015^{th}$ degree, let us generalize the problem for $n^{th}$ degree.

Consider the function:

$Q(x) = \sum_{i} \dfrac{r_i^k P(x)}{P'(r_i)(x-r_i)} - x^k$

where $P(x)$ is a $n^{th}$ degree polynomial with simple roots $r_1, r_2, r_3, \ldots , r_n$ and $k$ is any integer with $0 \leq k \leq (n-1)$ . Since $(x-r_i)|P(x)$ for all $i$ , the function $Q(x)$ is a polynomial of degree (at most) $n-1$ . For any root $r_j$ ,

$Q(r_j) = \sum_{i \neq j} \dfrac{r_i^k P(r_j)}{P'(r_i)(r_j - r_i)} + \lim_{x \to r_j} \dfrac{r_j^k P(x)}{P'(r_j)(x-r_j)} - r_j^k$ $= \lim_{x \to r_j} \left(\dfrac{r_j^k }{P'(r_j)} \right) \left( \dfrac{P(x) - p(r_j)}{x-r_j} \right) - r_j^k = \dfrac{r_j^k P'(r_j)}{P'(r_j)} - r_j^k = 0$

Therefore $Q(x) = 0$ for $n$ distinct values $r_1, r_2, \ldots, r_n$ , but since $Q(x)$ has a degree of at most $n-1$ , it must be identically zero. Thus:

$\sum_{i} \dfrac{r_i^k P(x)}{P'(r_i)(x-r_i)} - x^k = 0$

and, if $x \neq r_i$ for all $i$ , we have;

$\sum_{i} \dfrac{r_i^k}{P'(r_i)(x-r_i)} = \dfrac{x^k}{P(x)} \qquad ...(1)$

For $k=0$ in $(1)$ , we have:

$\sum_{i} \dfrac{1}{P'(r_i)(x-r_i)} = \dfrac{1}{P(x)}$

and by setting $x=0$ , we have $\displaystyle \sum_{i} \dfrac{1}{P'(r_i)r_i} = -\dfrac{1}{P(0)} = -\dfrac{1}{a_0}$

For $k=n-1$ in $(1)$ , we have:

$\sum_{i} \dfrac{r_i^{n-1}}{P'(r_i)(x-r_i)} = \dfrac{x^{n-1}}{P(x)} \quad \text{and} \quad \sum_{i} \left(\dfrac{r_i^{n-1}}{P'(r_i)} \right) \left(\dfrac{x}{x-r_i} \right) = \dfrac{x^n}{P(x)}$

By taking the infinite limit:

$\lim_{x \to \infty} \sum_{i} \left(\dfrac{r_i^{n-1}}{P'(r_i)} \right) \left(\dfrac{x}{x-r_i} \right) = \lim_{x \to \infty} \dfrac{x^n}{P(x)} \quad \text{and} \quad \sum_{i} \left(\dfrac{r_i^{n-1}}{P'(r_i)} \right) = \dfrac{1}{a_n}$

Additionally for $0 < k < n-2$ (that is $1 < k+1 < n-1$ ) in $(1)$ , we have:

$\sum_{i} \dfrac{r_i^{k+1}}{P'(r_i)(x-r_i)} = \dfrac{x^{k+1}}{P(x)}$

and for $x=0$ , we have, $\displaystyle -\sum_{i} \dfrac{r_i^{k+1}}{P'(r_i)(r_i)} = 0$ and $\displaystyle \sum_i \dfrac{r_i^k}{P'(r_i)} = 0$ .

We conclude that:

$\large{\sum_i \dfrac{r_i^k}{P'(r_i)} = \begin{cases} \color{#3D99F6}{-\dfrac{1}{a_0} \quad \text{ if } k=-1} \\ 0 \qquad \text{ if } 0 \leq k \leq n-2 \\ \color{#D61F06}{\dfrac{1}{a_n} \qquad \text{ if } k = n-1} \\ \end{cases} }$

We will sum upto the asked limit, and we also know that $a_z = \ln(e + z)$ . We have:

$\large{\sum_{k=0}^{2014} \left(\sum_{i=1}^{2015} \dfrac{r_i^{k-1}}{P'(r_i)} \right) = \dfrac{1}{a_{2015}} - \dfrac{1}{a_0} = \dfrac{1}{\ln(e+2015)} - \dfrac{1}{\ln(e)} \approx \boxed{-0.868}}$