Hola, A revisit to Beta, Gamma and Digamma

Let $J=\sum_{a=2}^{\infty} \left( \lim_{n\to \infty} \frac {g_n(f_n(a))}{a! }\right)$

And $I=\lim_{n\to \infty} g_n(f_n(a))$ $=\lim_{n\to \infty} \left(\int_0^1 (1-x^a)^n dx\right)^{1/n}$ $=\lim_{n\to \infty} \left(\frac {1}{a}\right) ^{1/n} \left(\frac {\Gamma (n+1)\Gamma (1/a)}{\Gamma \left( n+1+ \frac 1a\right)}\right) ^{1/n}$ $=\lim_{n\to \infty} \left(\frac {\Gamma (n+1)\Gamma (1/a)}{\Gamma \left( n+1+ \frac 1a\right)}\right) ^{1/n}$ $=\exp {\left( \lim_{n\to \infty} \frac {\ln( \Gamma (n+1))+\ln (\Gamma (1/a))-\ln\left(\Gamma\left(n+1+\frac 1a\right)\right)}{n}\right) }$

Applying L'Hospital rule we get $I=\exp {\left(\lim_{n\to \infty} \left(\psi(n+1)-\psi\left(n+1+\frac 1a\right)\right)\right)}$

But for large $x$ We have $\psi(x)\sim \ln (x)$

Hence we get $I=\exp {(\lim_{n\to \infty} (\ln (n+1)-\ln (n+1 + 1/a)))}=1$

Hence $J=\sum_{a=2}^{\infty} \frac {1}{a!} =e-2$