Find the sum of all the digits of the least positive integer $n$ such that $(P_2 * P_3 * $ ... $ * P_n)$ exceeds $2010$ .

Question

Let $P_k = 1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3}$ , where $k$ is a positive integer. Find the sum of all the digits of the least positive integer $n$ such that $(P_2 * P_3 * $ ... $ * P_n)$ exceeds $2010$ .

What I Tried: I have that :- $$\rightarrow P_k = 1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3} = \frac{(k - 1)(k + 1)^2}{k^3}$$ But I could not find way to use this. I thought this would give me some way to work on the problem but no.

Basically I have to find the least $n$ such that :- $$\displaystyle\prod_{k = 2}^{n} \bigg(1 + \frac{1}{k} - \frac{1}{k^2} - \frac{1}{k^3}\bigg) > 2010$$

I have no idea on how to find $n$ . I used Wolfram Alpha mainly for Trial and Error and to get the solution, and I found $n = 8038$ .
Check this here:- https://www.wolframalpha.com/input/?i=Product+k+%3D+2+to+8038%2C++%281+%2B+1%2Fk+-+1%2Fk%5E2+-+1%2Fk%5E3%29

So the final answer would be $19$ , but mathematically, how can you find $n$?

Can anyone help me?

Answer 1

Hint You already observed the key property

$$ P_k = \frac{(k - 1)(k + 1)^2}{k^3}$$

Therefore $$P_2*P_3 *P_4*...*P_n= \frac{1(3)^2}{2^3} \frac{2(4)^2}{3^3} \frac{3(5)^2}{4^3} ... \frac{(n - 1)(n + 1)^2}{n^3}$$

Now, just cancel as much as you can, what do you have leftover?

Answer 2

You can rewrite the product as telescoping product:

$$\prod_{n=2}^n P_k= \prod_{n=2}^n \left(\frac{k+1}{k}\right)^2\cdot \prod_{n=2}^n \frac{k-1}k = \frac{(n+1)^2}4\cdot \frac 1n >2010$$

$$\Leftrightarrow n+2+\frac 1n >8040$$

I think, now you can take it from there.