I haven't seen it before 2

By Maclaurin series , we have:

$\begin{aligned} \ln \left(\frac{1+x}{1-x}\right) & = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \frac{2x^7}{7} + ... \\ \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) & = x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + ... \\ \implies \sum_{n=2}^{99} 2 \mathfrak P_n & = \sum_{n=2}^{99} \ln \left(\frac{1+\frac{1}{n}}{1-\frac{1}{n}}\right) \\ & = \sum_{n=2}^{99} \ln \left(\frac{n+1}{n-1}\right) \\ & = \sum_{n=2}^{99} \left(\ln(n+1)-\ln(n-1)\right) \\ & = \sum_{n=3}^{100} \ln n - \sum_{n=1}^{98} \ln n \\ & = \ln 99 + \ln 100 - \ln 1 - \ln 2 \\ & = \ln \left(\frac{9900}{2}\right) = \ln 4950 \end{aligned}$

$\implies \alpha = \boxed{4950}$

Write given series as integral from 0 to 1/n of (1+x^2 + x^4............)dx

Use formula of infinite Geometric progression as x<1 and then integrate to get the sum as 1/2 [ ln(n+1)-ln(n-1) ]

Then to evaluate final answer use telescoping product

I integrated -(1/n^2 + 1/n^4 + 1/n^6 +...) and took the constant of integration as zero due to cognitive bias, so that I can arrive at the mentioned expression. Do I need to worry about the constant of integration here?

python code:

    import cmath as m
    def fun(x):
        return ((m.log(1+(1/x)))-(m.log(1-(1/x))))/2

    sum=0
    for i in range(2,100):
        sum=sum+2*fun(i)

    print(str(m.exp(sum.real)))