Question: Let $f_0(x)=\ln x$ and $f_n(x)=\int_0^xf_{n-1}(t)dt.$ Show that $$\lim_{n\to\infty}\frac{n!f_n(1)}{\ln n}=-1.$$
Solution: We have $f_0(x)=\ln x$ and $f_n(x)=\int_0^xf_{n-1}(t)dt, \forall n\in\mathbb{N}.$ Thus $$f_1(x)=\int_0^xf_0(t)dt=\int_0^x\ln(t)dt=x(\ln x-1)-\lim_{l\to 0+}l(\ln l-1)=x(\ln x-1)=\frac{x^1}{1!}\left(\ln x-\sum_{i=1}^1\frac{1}{i}\right).$$
Again $$f_2(x)=\int_0^xf_1(t)dt=\int_0^xt(\ln t-1)=\frac{x^2}{2}\ln x-\frac{x^2}{2}-\frac{x^2}{4}=\frac{x^2}{2}\left(\ln x-1-\frac{1}{2}\right)=\frac{x^2}{2!}\left(\ln x-\sum_{i=1}^2\frac{1}{i}\right).$$
By induction we can show that $$f_n(x)=\frac{x^n}{n!}\left(\ln x-\sum_{i=1}^n\frac{1}{i}\right),\forall n\in\mathbb{N}.$$
Hence, we can conclude that $$f_n(1)=-\frac{1}{n!}\sum_{i=1}^n\frac{1}{i},\forall n\in\mathbb{N}.$$
Let $\{a_n\}_{n\ge 2}$, be such that $$a_n=\frac{n!f_n(1)}{\ln n}, \forall n\ge 2.$$
Hence, $$a_n=\frac{n!}{\ln n}f_n(1)=\frac{n!}{\ln n}\left(-\frac{1}{n!}\sum_{i=1}^n\frac{1}{i}\right)=-\frac{1}{\ln n}\sum_{i=1}^n\frac{1}{i}, \forall n\ge 2.$$
Now let $\{b_n\}_{n\ge 2}$, be such that $$b_n=\sum_{i=1}^n\frac{1}{i}-\ln n, \forall n\ge 2.$$
We know that $\{b_n\}_{n\ge 2}$ is a convergent sequence, i.e., it has a finite limit. For the proof of the same, refer to this link: For $n\ge 1$, define $a_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n$. Prove that $\lim_{n\to\infty} a_n$ exists. Of course this limit is the Euler-Mascheroni constant $\gamma$.
Again let $\{c_n\}_{n\ge 2}$, be such that $$c_n=\frac{b_n}{\ln n}+1=\frac{1}{\ln n}\sum_{i=1}^n\frac{1}{i}, \forall n\ge 2.$$
Clearly, $\{c_n\}_{n\ge 2}$ is a convergent sequence and $$\lim_{n\to\infty}c_n=\gamma.0+1=1.$$
Observe that $a_n=-c_n, \forall n\ge 2$, hence $\{a_n\}_{n\ge 2}$ is also a convergent sequence and we have $$\lim_{n\to\infty}a_n=-1.$$
Hence, we have $$\lim_{n\to\infty}\frac{n!f_n(1)}{\ln n}=-1.$$
Is this solution correct and rigorous enough? An alternate solution will always be appreciated.
