Find $S=\bigg\lfloor {\sum_{k=1}^{99} \frac{1}{\sqrt{r_k-1}}} \bigg \rfloor$ If $$ \prod_{k=1}^{99} (x+k) = \sum_{k=0}^{99} a_k \cdot x^k$$
Also $$ M= \frac{\sum_{k=1}^{99} k\cdot a_k}{100!}= \sum_{i=1}^{99} \frac{1}{r_i}$$
Where $r_i<r_j$, $∀$ $i<j$. Then find $S$ which is given by
$$S=\bigg\lfloor {\sum_{k=1}^{99} \frac{1}{\sqrt{r_k-1}}} \bigg \rfloor$$
My approach:
I don't know what's the role of $r_i<r_j$, $∀$ $i<j$.
So, I wrote $$\frac{1}{v_k}= \frac{k\cdot a_k}{100!}$$
Because,$${\sum_{k=1}^{99} \frac{1}{\sqrt{r_k-1}}}={\sum_{k=1}^{99} \frac{1}{\sqrt{v_k-1}}}$$
Which gave, $$\frac{1}{\sqrt{v_k-1}}= \sqrt{ \frac{k\cdot a_k}{100!-k\cdot 
 a_k}} $$
But now I am unable to solve this summation series. Could someone help me out?
 A: I suppose that the $r_i$ are integers such that $1<r_1<\ldots<r_{99}$.
The statement is very enigmatic, but I think that I have a solution.
Call $P$ the polynomial given by
$$P(x) = \prod_{k=1}^{99}(x+k) = \sum_{k=0}^{99} a_k x^k.$$
Then
$$P'(x) = \sum_{i=1}^{99}\prod_{k \ne i}(x+k) = \sum_{k=0}^{99}k a_k x^{k-1}.$$
By division, if $x \notin \{1,\ldots,99\}$,
$$\frac{P'(x)}{P(x)} = \sum_{i=1}^{99}\frac{1}{x+i} = \frac{\sum_{k=0}^{99}k a_k x^{k-1}}{\prod_{k=1}^{99}(x+k)}.$$
Applying these two equalities to $x=1$ yields
$$\frac{P'(1)}{P(1)} = \sum_{i=1}^{99}\frac{1}{1+i} = \frac{\sum_{k=0}^{99}k a_k}{100!}$$
The only choice of integers such that $1<r_1<\ldots<r_{99}$ and
$$\sum_{i=1}^{99}\frac{1}{r_i} = \frac{\sum_{k=0}^{99}k a_k}{100!}$$
are given by $r_i=i+1$ for every $i$. Indeed, the condition $1<r_1<\ldots<r_{99}$ forces $r_i \ge i+1$ for every $i$, and if one of this inequality was strict, we would get $\sum_{i=1}^{99}\frac{1}{r_i} < \frac{\sum_{k=0}^{99}k a_k}{100!}$.
Hence the quantity to be computed is $S = \lfloor \sum_{i=1}^{99}\frac{1}{\sqrt{i}}\rfloor$.
Since the map $x \mapsto x^{-1/2}$ is decreasing on $\mathbb{R}_+$, we have for every $i \ge 1$
$$\int_i^{i+1} x^{-1/2}dx < i^{-1/2} < \int_{i-1}^i x^{-1/2}dx.$$
By summation over $i \in \{2,\ldots,99\}$,
$$17 < 2\sqrt{100}-2\sqrt{2} = \int_2^{100} x^{-1/2}dx < \sum_{i=2}^{99} i^{-1/2} < \int_1^{99} x^{-1/2}dx = 2\sqrt{99}-2\sqrt{1} < 18,$$
so
$$18 < \sum_{i=1}^{99} i^{-1/2} < 19.$$
Hence $S=18$.
