I'm confused about the title #RMM (2)

My approach

For $a>0$ , defined $\begin{aligned} f(x) & = \dfrac{1}{\,(x^2+2xa +a^2 )+\,(x+a) }=\dfrac{1}{\,(x+a)^2+\,(x+a)} \\& = \dfrac{1}{\,(x+a+1)(x+a)}= \dfrac{1}{x+a}-\dfrac{1}{x+a+1}\end{aligned}$ Thus $\begin{aligned} f^{n}(x) & =\underbrace{\dfrac{d}{dx}\left(\cdots \dfrac{d}{dx}\left(\dfrac{d}{dx}\left(\dfrac{d}{dx}\left(\dfrac{1}{x+a} -\dfrac{1}{x+a +1}\right)\right)\right)\cdots \right)}_{\text{n times}}\\& = \dfrac{(-1)^n n!}{\,(x+a)^{n+1}}-\dfrac{(-1)^n n!}{\,(x+a+1)^{n+1} }=(-1)^n n!\left[\dfrac{1}{(x+a)^{n+1}}-\dfrac{1}{(x+a+1)^{n+1}}\right]\end{aligned}$ replacing $x$ by $k$ and thus $\left(\left| \lim_{p\to \infty} \sum_{k=1}^{p}f^{n} (k) \right|_{> 0}\right)^{\frac{1}{n^2}}=\lim_{p\to \infty} \left(\sum_{k=1}^{p} \left[\dfrac{n!}{(k+a)^{n+1}}-\dfrac{n!}{(k+a+1)^{n+1}}\right]\right)^{\frac{1}{n^2}}$ Since the partial sum of $\sum_{k=1}^{p}\left[\dfrac{n!}{(k+a)^{n+1}}-\dfrac{n!}{(k+a+1)^{n+1}}\right]= \dfrac{n!}{(a+1)^{n+1}}-\dfrac{n!}{(p+a+1)^{n+1}}$ As $p\to \infty$ and hence the $\lim_{p\to \infty} \sum_{k=1}^{p}\left[\dfrac{n!}{(k+a)^{n+1}}-\dfrac{n!}{(k+a+1)^{n+1}}\right] =\dfrac{n!}{(a+1)^{n+1}}-0$ Finally we obtain that $\begin{aligned} L& = \left(\lim_{n\to \infty} \dfrac{ n!}{(a+1)^{n+1}}\right)^{\frac{1}{n^2}}\sim \left(\lim_{n\to \infty} \dfrac{\sqrt{2\pi n}\ }{(a+1)^{n+1}}\left(\dfrac{n}{e}\right)^n \right)^{\frac{1}{n^2}} \\& = \lim_{n\to \infty} \dfrac{\,(2\pi)^{\frac{1}{2n^2}}\cdot n^{\frac{1+2n}{n^2}}}{e^{\frac{n}{n^2}}\,(a+1)^{\frac{n+1}{n^2}}}= \lim_{n\to \infty} \exp\left(\dfrac{(1+2n)\log n}{n^2}\right)=e^0 =1 \end{aligned}$

Relevant wiki: Stirling's Formula

$\begin{aligned} f(x) & = \frac 1{x^2+(2a+1)x+a^2+a} = \frac 1{x^2 + 2ax+a^2 + x + a} = \frac 1{(x+a)^2 + x + a} = \frac 1{(x+a)(x+a+1)} = \frac 1{x+a} - \frac 1{x+a+1} \end{aligned}$

$\begin{aligned} \implies f^{(n)} (x) & = (-1)^n n! \left(\frac 1{(x+a)^{n+1}} - \frac 1{(x+a+1)^{n+1}} \right) \\ \sum_{k=1}^\infty f^{(n)}(k) & = \sum_{k=1}^\infty (-1)^n n! \left(\frac 1{(k+a)^{n+1}} - \frac 1{(k+a+1)^{n+1}} \right) = \frac {(-1)^n n!}{(1+a)^{n+1}} \end{aligned}$

Therefore,

$\begin{aligned} L & = \lim_{n \to \infty} \sqrt[n^2]{\left|\sum_{k=1}^\infty f^{(n)}(k) \right|} = \lim_{n \to \infty} \left(\frac {\color{#3D99F6}n!}{(1+a)^{n+1}}\right)^{\frac 1{n^2}} & \small \color{#3D99F6} \text{By Stirling's formula} \\ & = \lim_{n \to \infty} \left(\frac {\color{#3D99F6}\sqrt{2\pi n}n^ne^{-n}}{(1+a)^{n+1}}\right)^{\frac 1{n^2}} = \lim_{n \to \infty} \frac {(2\pi)^{\frac 1{2n^2}}n^{\frac 1n+\frac 1{2n^2}}}{e^\frac 1n(1+a)^{\frac 1n+\frac 1{n^2}}} \\ & = \boxed 1 \end{aligned}$