For $n \in \Bbb N$, let $g_n(x)=\displaystyle x^{\frac{1}{n}}$ and $f_n(a)=\displaystyle \int_0^1(1-x^a)^n dx$. For $g$ and $f$ defined above, if $$\sum_{a=2}^\infty\left( \lim_{n \to \infty} \frac{g_n(f_n(a))}{a!} \right)=\alpha$$ then find $\alpha$.
I tried solving the limit by taking logarithm and then applying L'hopital's rule, but it becomes even more complicated.
Clearly, when $a>0$, $0\leq 1-x^a\leq 1$ for $x\in[0,1]$ and $\max_{x\in[0,1]}(1-x^a)=1-0=1$. So $\left(\displaystyle \int_0^1(1-x^a)^n\,dx\right)^{\frac1n}\leq 1.$
Since $1-x^a$ is continuous, for every $0<\epsilon<1$ we can find $\delta>0$ such that $1-\epsilon\leq 1-x^a$ for $x\in[0,\delta]$, hence $$\left(\int_0^1(1-x^a)^n\,dx\right)^{\frac1n}\geq\left(\int_0^\delta(1-x^a)^n\,dx\right)^{\frac1n}\geq (1-\epsilon)\delta^{\frac 1n}.$$ Hence $$\lim_{n\to\infty}\left(\int_0^1(1-x^a)^n\,dx\right)^{\frac1n}=1$$ and $$\alpha=\sum_{a=2}^\infty\left(\lim_{n\to\infty}\frac{g_n(f_n(a))}{a!}\right)=\sum_{a=2}^\infty\frac{1}{a!}=e-2.$$