Cfrgtkky

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:

$$ frac{(epsilon^2)!}{s ((2epsilon-2)!)^2(2epsilon-1 )} frac{(epsilon-1)!^2}{(epsilon-1)^2!} = 1 + O(1)- epsilon^2 ( gamma + 2ln(epsilon-1) - B(epsilon-1) ) $$

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:

$$ frac{(s^2)!}{(s!)^2} = frac{s^2cdot(s^2-1)cdot(s^2-2)cdots(s+1)}{s!} $$

Now, we can do some further cancellation using the identity $s^2 - r^2= (s-r)(s+r)$:

$$ frac{(s^2)!}{(s!)^2} = {scdot(s+1)cdot(s^2-2)cdots(s+1)} = s(s+1)^2(s+2)^2cdots(s^2 -2)(s^2-3)dots $$

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:

$$sum_{i}^{s^2 -3s +2} frac{1}{m_i} = sum_r^{(s-1)^2} frac{1}{r} - sum_{r=1}^{s-1} frac{1}{r^2} = gamma + 2ln(s-1) +O(frac{1}{s^2}) - B(s-1)$$

where $B(s-1) = sum_{r=1}^{s-1} frac{1}{r^2}$. Now using the above as a substitution:

$$ frac{(s^2)!}{s ((2s-2)!)^2(2s-1 )} frac{(s-1)!^2}{(s-1)^2!} = 1 + O(1)- s^2 ( gamma + 2ln(s-1) - B(s-1) ) + dots $$

Now let us put $s to epsilon$ where $epsilon$ is a nil potent matrix such that $epsilon^4 = 0$. Hence, the expression becomes:

$$ frac{(epsilon^2)!}{s ((2epsilon-2)!)^2(2epsilon-1 )} frac{(epsilon-1)!^2}{(epsilon-1)^2!} = 1 + O(1)- epsilon^2 ( gamma + 2ln(epsilon-1) - B(epsilon-1) ) $$

Note the $1$ used above is actually the identity. Using this it should be possible to find $B(epsilon)$'s first two terms.