Question
What is the taylor expansion of $\Gamma(n+1)$ of the first $4$ terms? Are the results below correct?
I thought of an interesting way to calculate the following quantity:
$$ \prod_{(s-1)^2> m \neq I^2} (s^2 -m) =\frac{(s^2)!}{s ((2s-2)!)^2(2s-1 )} $$
Where $m$ is an integer which is not a square. Further using this I realised I could calculate partial sums of the Basel problem:
$\epsilon$ is a nil potent matrix such that $\epsilon^4 = 0$. Hence, the expression becomes:
$$ \frac{(\epsilon^2)!}{s ((2\epsilon-2)!)^2(2\epsilon-1 )} \frac{(\epsilon-1)!^2}{(\epsilon-1)^2!} = 1 + O(1)- \epsilon^2 ( \gamma + 2\ln(\epsilon-1) - B(\epsilon-1) ) $$
Note the $1$ used above is actually the identity. Using this it should be possible to find $B(\epsilon)$'s first two terms.
Background and Motivation
I was pondering about the following:
$$ \frac{(s^2)!}{(s!)^2} = \frac{s^2\cdot(s^2-1)\cdot(s^2-2)\cdots(s+1)}{s!} $$
Now, we can do some further cancellation using the identity $s^2 - r^2= (s-r)(s+r)$:
$$ \frac{(s^2)!}{(s!)^2} = {s\cdot(s+1)\cdot(s^2-2)\cdots(s+1)} = s(s+1)^2(s+2)^2\cdots(s^2 -2)(s^2-3)\dots $$
Hence, we can the above expression properly:
$$ \frac{(s^2)!}{(s!)^2} = s (s^2 -(s-1)^2))\prod_{1 \leq k \leq s-2} (s+k)^2 \prod_{(s-1)^2> m \neq I^2 } (s^2 -m)$$
Where $I$ is an arbitrary integer and now re-writing the above expression:
$$ \frac{(s^2)!}{s ((2s-2)!)^2(2s-1 )} = \prod_{(s-1)^2> m \neq I^2} (s^2 -m)$$
Now since there are $(s-1)^2 - (s-1) = s^2 -3s +2$ non-square numbers less than $(s-1)^2$:
$$ \frac{(s^2)!}{s ((2s-2)!)^2(2s-1 )} = \prod_{i=1}^{s^2 -3s +2} (s^2 -m_i)$$
Where $m_i$ is the $i$'th non-square number. Taking $m_i$ common from the R.H.S using $\prod_{i} m_i = \frac{(s-1)^2!}{(s-1)!^2}$:
$$ \frac{(s^2)!}{s ((2s-2)!)^2(2s-1 )} = \frac{(s-1)^2!}{(s-1)!^2}\prod_{i=1}^{s^2 -3s +2} (\frac{s^2}{m_i} - 1) $$
Expanding out the R.H.S from smallest term to largest:
$$ \frac{(s^2)!}{s ((2s-2)!)^2(2s-1 )} \frac{(s-1)!^2}{(s-1)^2!} = 1 - s^2(\sum_{i=1}^{s^2 -3s +2} \frac{1}{m_i}) + \dots + \prod_{i=1}^{s^2 -3s +2} (\frac{s^2}{m_i} ) $$
Now we use the identity:
$$\sum_{i}^{s^2 -3s +2} \frac{1}{m_i} = \sum_r^{(s-1)^2} \frac{1}{r} - \sum_{r=1}^{s-1} \frac{1}{r^2} = \gamma + 2\ln(s-1) +O(\frac{1}{s^2}) - B(s-1)$$
where $B(s-1) = \sum_{r=1}^{s-1} \frac{1}{r^2}$. Now using the above as a substitution:
$$ \frac{(s^2)!}{s ((2s-2)!)^2(2s-1 )} \frac{(s-1)!^2}{(s-1)^2!} = 1 + O(1)- s^2 ( \gamma + 2\ln(s-1) - B(s-1) ) + \dots $$
Now let us put $s \to \epsilon$ where $\epsilon$ is a nil potent matrix such that $\epsilon^4 = 0$. Hence, the expression becomes:
$$ \frac{(\epsilon^2)!}{s ((2\epsilon-2)!)^2(2\epsilon-1 )} \frac{(\epsilon-1)!^2}{(\epsilon-1)^2!} = 1 + O(1)- \epsilon^2 ( \gamma + 2\ln(\epsilon-1) - B(\epsilon-1) ) $$
Note the $1$ used above is actually the identity. Using this it should be possible to find $B(\epsilon)$'s first two terms.