Find $\lim_{n \rightarrow \infty} \int_{0}^{x} \frac{(-\ln{z})^n}{n!} dz$ Find $\lim_{n \rightarrow \infty} \int_{0}^{x} \frac{(-\ln{z})^n}{n!} dz$ where $x \in (0,1)$ .
I was thinking of using Dominated Convergence Theorem to the function $f_n(z)=\frac{(-\ln{z})^n}{n!} dz$
But couldn't find its limiting function.
Any other approach which can help?
 A: Let us first consider the integral, $${I(n)} = \int_0^{x}\frac{(-\ln z)^n}{n!}dz$$
Using Integration by Parts,
$$I(n) = \frac{z(-\ln z)^n}{n!}|_0^{x}+n\int_0^{x}\frac{(-\ln z)^{n-1}}{n!}dz=\frac{x(-\ln x)^n}{n!}+I(n-1)$$
If you observe, the above is a telescopic sum,
$$\therefore I(n)=I(0)+x\sum_{i=1}^{n}\frac{(-\ln x)^i}{i!}$$
Since $I(0) =x$,
$$I(n)=x\sum_{i=0}^{n}\frac{(-\ln x)^i}{i!}$$
As $n \to \infty$,
$$\lim_{n \to \infty}I(n)=x\sum_{i=0}^{\infty}\frac{(-\ln x)^i}{i! }=x\exp{(-\ln x)}=1$$
A: Enforcing the substitution $z\mapsto e^{-z}$ and recalling that $n!=\Gamma(n+1)=\int_0^\infty z^ne^{-z}\,dz$ we can write
$$\begin{align}
f_n(x)&=\int_0^x\frac{(-\log(z))^n}{n!}\,dz\\\\
&=\frac{1}{n!}\int_{\log(1/x)}^\infty z^ne^{-z}\,dz\\\\
&=\frac1{n!}\int_0^\infty z^ne^{-z}\,dz-\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz\\\\
&=1-\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz
\end{align}$$
Finally, using the estimate 
$$\left|\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz\right|\le \frac{\log^{n+1}(1/x)}{n!} $$
the squeeze theorem guarantees that $\lim_{n\to \infty }\frac1{n!}\int_0^{\log(1/x)}z^ne^{-z}\,dz=0$.
Putting it all together, yields the coveted limit
$$\lim_{n\to\infty}f_n(x)=1$$
for $x\in (0,1)$.  And we are done!
A: Not a direct proof but one can note that $$f(z)=\frac{(-\ln z)^n}{n!}1_{0<z<1}$$ is the density function of $\prod\limits_{i=1}^{n+1} X_i$ where $X_1,\ldots,X_{n+1}$ are i.i.d uniform on $(0,1)$.
So for $x\in (0,1)$,
\begin{align}
\int_0^x f(z)\,dz&=P\left(\prod_{i=1}^{n+1} X_i \le x\right)
\\&=P\left(\sum_{i=1}^{n+1} \ln X_i \le \ln x\right)
\end{align}
Now $-\ln X_i$'s are i.i.d exponential with mean $1$, so by classical CLT
$$\sqrt{n+1}\left(\frac1{n+1}\sum_{i=1}^{n+1} \ln X_i +1\right)\stackrel{L}\longrightarrow N(0,1)$$
Hence, 
$$
P\left(\sum_{i=1}^{n+1} \ln X_i \le \ln x\right)\approx \Phi\left(\frac{\ln x}{\sqrt{n+1}}+\sqrt{n+1}\right) \stackrel{n\to\infty}\longrightarrow \Phi(\infty)=1
$$
