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For each nonnegative integer $n$ we have $$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$ $$(\pi-S_n)\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)}p_k(\ln\pi).$$

As far as I see, $P_n(x)$ from the question equals $\widetilde P_n(x)$ from the Wikipedia article. Next, as I see, $e^xp_k(x)=\int_{0}^{x} y^ke^y dy+p_k(0)$ (by the way, in the question is written that $p_k(0)=k!$, whereas Mathcad calculations suggest that $p_k(0)=(-1)^kk!$; for instance, I obtained $\int x^3e^x dx=(x^3-3x^2-6x-6)e^x+C$ and $\int x^4e^x dx=(x^4-4x^3+12x^2-24x+24)e^x+C$).

Thus we have $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(\ln\pi)=$$ $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}e^{-\ln\pi}\left(\int_{0}^{\ln\pi} y^ke^y dy+p_k(0)\right)=$$ $$e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}\int_{0}^{\ln\pi} y^ke^y dy+e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)=$$ $$\frac 1\pi I_n+\frac 1\pi\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0).$$

Then $$I_n-J_n=\left(1-\frac{S_n}{\pi}\right)\left(I_n+\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)\right).$$

Thus we have to bound the absolute value of $T_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)$. I assume that $p_k(0)=(-1)^kk!$. Then $$T_n= (-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot {n\choose k}{n+k\choose k}k!=$$ $$(-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{n!}{(n-k)!k!}\cdot\frac{(n+k)!}{n!k!}\cdot k!=$$ $$(-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{(n+k)!}{(n-k)!k!}.$$

A rather rough estimation is $(n+k)!\le (2n)!$ for each nonegative integer $k\le n$. Thus

$$|T_n|\le \sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{(2n)!}{(n-k)!k!}=$$ $$\frac{(2n)!}{n!}\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{n!}{(n-k)!k!}=\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$

Unfortunately, I expect that we cannot compensate the growth of this bound by suitable choice of $f(n)$. Indeed, approximating $|S_n-\pi|$ by the last summand $$\frac{4}{16^k(8k+1)} -\frac{2}{16^k(8k+4)} -\frac{1}{16^k(8k+5)} -\frac{1}{16^k(8k+6)}$$ in the sum for $S_n$, we obtain that $$|S_n-\pi|\approx \frac{120k^2+151k+47}{(8k+1)(8k+4)(8k+5)(8k+6)}\approx \frac{15}{64}16^{-k}k^{-2},$$ where $k=f(n)$. This value should compensate the value $$\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$ Neglecting the factors of the smaller growth, we obtain that $16^{f(n)}\ge\approx \frac{(2n)!}{n!}$. But then $$\lim_{n\to\infty} \frac{16^{f(n)}e^{8f(n)}}{n!}\ge\approx e^{8f(n)}\frac{(2n!)}{n!^2}=\lim_{n\to\infty} e^{8f(n)}{2n\choose n}=\infty,$$ which violates $(2)$.

For each nonnegative integer $n$ we have $$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$ $$(\pi-S_n)\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)}p_k(\ln\pi).$$

As far as I see, $P_n(x)$ from the question equals $\widetilde P_n(x)$ from the Wikipedia article. Next, as I see, $e^xp_k(x)=\int_{0}^{x} y^ke^y dy+p_k(0)$ (by the way, in the question is written that $p_k(0)=k!$, whereas Mathcad calculations suggest that $p_k(0)=(-1)^kk!$; for instance, I obtained $\int x^3e^x dx=(x^3-3x^2-6x-6)e^x+C$ and $\int x^4e^x dx=(x^4-4x^3+12x^2-24x+24)e^x+C$).

Thus we have $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(\ln\pi)=$$ $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}e^{-\ln\pi}\left(\int_{0}^{\ln\pi} y^ke^y dy+p_k(0)\right)=$$ $$e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}\int_{0}^{\ln\pi} y^ke^y dy+e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)=$$ $$\frac 1\pi I_n+\frac 1\pi\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0).$$

Then $$I_n-J_n=\left(1-\frac{S_n}{\pi}\right)\left(I_n+\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)\right).$$

Thus we have to bound the absolute value of $T_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)$. I assume that $p_k(0)=(-1)^kk!$. Then $$T_n= (-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot {n\choose k}{n+k\choose k}k!=$$ $$(-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{n!}{(n-k)!k!}\cdot\frac{(n+k)!}{n!k!}\cdot k!=$$ $$(-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{(n+k)!}{(n-k)!k!}.$$

A rather rough estimation is $(n+k)!\le (2n)!$ for each nonegative integer $k\le n$. Thus

$$|T_n|\le \sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{(2n)!}{(n-k)!k!}=$$ $$\frac{(2n)!}{n!}\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{n!}{(n-k)!k!}=\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$

Unfortunately, I expect that we cannot compensate the growth even of the last summand $(-1)^n\frac 1{\ln^{n+1} \pi}\cdot \frac{(2n)!}{0!n!}$ for $T_n$ of this bound by suitable choice of $f(n)$. Indeed, approximating $|S_n-\pi|$ by the last summand $$\frac{4}{16^k(8k+1)} -\frac{2}{16^k(8k+4)} -\frac{1}{16^k(8k+5)} -\frac{1}{16^k(8k+6)}$$ in the sum for $S_n$, we obtain that $$|S_n-\pi|\approx \frac{120k^2+151k+47}{(8k+1)(8k+4)(8k+5)(8k+6)}\approx \frac{15}{64}16^{-k}k^{-2},$$ where $k=f(n)$. This value should compensate the value $$\frac 1{\ln^{n+1} \pi}\cdot \frac{(2n)!}{0!n!}.$$ Neglecting the factors of the smaller growth, we obtain that $16^{f(n)}\ge\approx \frac{(2n)!}{n!}$. But then $$\lim_{n\to\infty} \frac{16^{f(n)}e^{8f(n)}}{n!}\ge\approx e^{8f(n)}\frac{(2n!)}{n!^2}=\lim_{n\to\infty} e^{8f(n)}{2n\choose n}=\infty,$$ which violates $(2)$.

For each nonnegative integer $n$ we have $$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$ $$(\pi-S_n)\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)}p_k(\ln\pi).$$

As far as I see, $P_n(x)$ from the question equals $\widetilde P_n(x)$ from the Wikipedia article. Next, as I see, $e^xp_k(x)=\int_{0}^{x} y^ke^y dy+p_k(0)$ (by the way, in the question is written that $p_k(0)=k!$, whereas Mathcad calculations suggest that $p_k(0)=(-1)^kk!$; for instance, I obtained $\int x^3e^x dx=(x^3-3x^2-6x-6)e^x+C$ and $\int x^4e^x dx=(x^4-4x^3+12x^2-24x+24)e^x+C$).

Thus we have $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(\ln\pi)=$$ $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}e^{-\ln\pi}\left(\int_{0}^{\ln\pi} y^ke^y dy+p_k(0)\right)=$$ $$e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}\int_{0}^{\ln\pi} y^ke^y dy+e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)=$$ $$\frac 1\pi I_n+\frac 1\pi\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0).$$

Then $$I_n-J_n=\left(1-\frac{S_n}{\pi}\right)\left(I_n+\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)\right).$$

Thus we have to bound the absolute value of $T_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)$. I assume that $p_k(0)=(-1)^kk!$. Then $$T_n= (-1)^n\sum_{k=0}^n \ln^{-k-1} \pi {n\choose k}{n+k\choose k}k!=$$ $$(-1)^n\sum_{k=0}^n \ln^{-k-1} \pi \frac{n!}{(n-k)!k!}\cdot\frac{(n+k)!}{n!k!}\cdot k!=$$ $$(-1)^n\sum_{k=0}^n \ln^{-k-1} \pi \frac{(n+k)!}{(n-k)!k!}.$$

A rather rough estimation is $(n+k)!\le (2n)!$ for each nonegative integer $k\le n$. Thus

$$|T_n|\le \sum_{k=0}^n \ln^{-k-1} \pi \frac{(2n)!}{(n-k)!k!}=$$ $$\frac{(2n)!}{n!}\sum_{k=0}^n \ln^{-k-1} \pi \frac{n!}{(n-k)!k!}=\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$

Unfortunately, I expect that we cannot compensate the growth of this bound by suitable choice of $f(n)$. Indeed, approximating $|S_n-\pi|$ by the last summand $$\frac{4}{16^k(8k+1)} -\frac{2}{16^k(8k+4)} -\frac{1}{16^k(8k+5)} -\frac{1}{16^k(8k+6)}$$ in the sum for $S_n$, we obtain that $$|S_n-\pi|\approx \frac{120k^2+151k+47}{(8k+1)(8k+4)(8k+5)(8k+6)}\approx \frac{15}{64}16^{-k}k^{-2},$$ where $k=f(n)$. This value should compensate the value $$\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$ Neglecting the factors of the smaller growth, we obtain that $16^{f(n)}\ge\approx \frac{(2n)!}{n!}$. But then $$\lim_{n\to\infty} \frac{16^{f(n)}e^{8f(n)}}{n!}\ge\approx e^{8f(n)}\frac{(2n!)}{n!^2}=\lim_{n\to\infty} e^{8f(n)}{2n\choose n}=\infty,$$ which violates $(2)$.

For each nonnegative integer $n$ we have $$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$ $$(\pi-S_n)\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)}p_k(\ln\pi).$$

As far as I see, $P_n(x)$ from the question equals $\widetilde P_n(x)$ from the Wikipedia article. Next, as I see, $e^xp_k(x)=\int_{0}^{x} y^ke^y dy+p_k(0)$ (by the way, in the question is written that $p_k(0)=k!$, whereas Mathcad calculations suggest that $p_k(0)=(-1)^kk!$; for instance, I obtained $\int x^3e^x dx=(x^3-3x^2-6x-6)e^x+C$ and $\int x^4e^x dx=(x^4-4x^3+12x^2-24x+24)e^x+C$).

Thus we have $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(\ln\pi)=$$ $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}e^{-\ln\pi}\left(\int_{0}^{\ln\pi} y^ke^y dy+p_k(0)\right)=$$ $$e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}\int_{0}^{\ln\pi} y^ke^y dy+e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)=$$ $$\frac 1\pi I_n+\frac 1\pi\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0).$$

Then $$I_n-J_n=\left(1-\frac{S_n}{\pi}\right)\left(I_n+\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)\right).$$

Thus we have to bound the absolute value of $T_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)$. I assume that $p_k(0)=(-1)^kk!$. Then $$T_n= (-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot {n\choose k}{n+k\choose k}k!=$$ $$(-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{n!}{(n-k)!k!}\cdot\frac{(n+k)!}{n!k!}\cdot k!=$$ $$(-1)^n\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{(n+k)!}{(n-k)!k!}.$$

A rather rough estimation is $(n+k)!\le (2n)!$ for each nonegative integer $k\le n$. Thus

$$|T_n|\le \sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{(2n)!}{(n-k)!k!}=$$ $$\frac{(2n)!}{n!}\sum_{k=0}^n \frac 1{\ln^{k+1} \pi}\cdot \frac{n!}{(n-k)!k!}=\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$

Unfortunately, I expect that we cannot compensate the growth of this bound by suitable choice of $f(n)$. Indeed, approximating $|S_n-\pi|$ by the last summand $$\frac{4}{16^k(8k+1)} -\frac{2}{16^k(8k+4)} -\frac{1}{16^k(8k+5)} -\frac{1}{16^k(8k+6)}$$ in the sum for $S_n$, we obtain that $$|S_n-\pi|\approx \frac{120k^2+151k+47}{(8k+1)(8k+4)(8k+5)(8k+6)}\approx \frac{15}{64}16^{-k}k^{-2},$$ where $k=f(n)$. This value should compensate the value $$\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$ Neglecting the factors of the smaller growth, we obtain that $16^{f(n)}\ge\approx \frac{(2n)!}{n!}$. But then $$\lim_{n\to\infty} \frac{16^{f(n)}e^{8f(n)}}{n!}\ge\approx e^{8f(n)}\frac{(2n!)}{n!^2}=\lim_{n\to\infty} e^{8f(n)}{2n\choose n}=\infty,$$ which violates $(2)$.

This answer is unfinished.

For each nonnegative integer $n$ we have $$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$ $$(\pi-S_n)\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)}p_k(\ln\pi).$$

As far as I see, $P_n(x)$ from the question equals $\widetilde P_n(x)$ from the Wikipedia article. Next, as I see, $e^xp_k(x)=\int_{0}^{x} y^ke^y dy+p_k(0)$ (by the way, in the question is written that $p_k(0)=k!$, whereas Mathcad calculations suggest that $p_k(0)=(-1)^kk!$; for instance, I obtained $\int x^3e^x dx=(x^3-3x^2-6x-6)e^x+C$ and $\int x^4e^x dx=(x^4-4x^3+12x^2-24x+24)e^x+C$).

Thus we have $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(\ln\pi)=$$ $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}e^{-\ln\pi}\left(\int_{0}^{\ln\pi} y^ke^y dy+p_k(0)\right)=$$ $$e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}\int_{0}^{\ln\pi} y^ke^y dy+e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)=$$ $$\frac 1\pi I_n+\frac 1\pi\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0).$$

Then $$I_n-J_n=\left(1-\frac{S_n}{\pi}\right)\left(I_n+\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)\right).$$

Thus we have to bound the absolute value of $T_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)$. I assume that $p_k(0)=(-1)^kk!$. Then $$T_n= (-1)^n\sum_{k=0}^n \ln^{-k-1} \pi {n\choose k}{n+k\choose k}k!=$$ $$(-1)^n\sum_{k=0}^n \ln^{-k-1} \pi \frac{n!}{(n-k)!k!}\cdot\frac{(n+k)!}{n!k!}\cdot k!=$$ $$(-1)^n\sum_{k=0}^n \ln^{-k-1} \pi \frac{(n+k)!}{(n-k)!k!}.$$

A rather rough estimation is $(n+k)!\le (2n)!$ for each nonegative integer $k\le n$. Thus

$$|T_n|\le \sum_{k=0}^n \ln^{-k-1} \pi \frac{(2n)!}{(n-k)!k!}=$$ $$\frac{(2n)!}{n!}\sum_{k=0}^n \ln^{-k-1} \pi \frac{n!}{(n-k)!k!}=\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$

For each nonnegative integer $n$ we have $$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$ $$(\pi-S_n)\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)}p_k(\ln\pi).$$

As far as I see, $P_n(x)$ from the question equals $\widetilde P_n(x)$ from the Wikipedia article. Next, as I see, $e^xp_k(x)=\int_{0}^{x} y^ke^y dy+p_k(0)$ (by the way, in the question is written that $p_k(0)=k!$, whereas Mathcad calculations suggest that $p_k(0)=(-1)^kk!$; for instance, I obtained $\int x^3e^x dx=(x^3-3x^2-6x-6)e^x+C$ and $\int x^4e^x dx=(x^4-4x^3+12x^2-24x+24)e^x+C$).

Thus we have $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(\ln\pi)=$$ $$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}e^{-\ln\pi}\left(\int_{0}^{\ln\pi} y^ke^y dy+p_k(0)\right)=$$ $$e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}\int_{0}^{\ln\pi} y^ke^y dy+e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)=$$ $$\frac 1\pi I_n+\frac 1\pi\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0).$$

Then $$I_n-J_n=\left(1-\frac{S_n}{\pi}\right)\left(I_n+\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)\right).$$

Thus we have to bound the absolute value of $T_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)$. I assume that $p_k(0)=(-1)^kk!$. Then $$T_n= (-1)^n\sum_{k=0}^n \ln^{-k-1} \pi {n\choose k}{n+k\choose k}k!=$$ $$(-1)^n\sum_{k=0}^n \ln^{-k-1} \pi \frac{n!}{(n-k)!k!}\cdot\frac{(n+k)!}{n!k!}\cdot k!=$$ $$(-1)^n\sum_{k=0}^n \ln^{-k-1} \pi \frac{(n+k)!}{(n-k)!k!}.$$

A rather rough estimation is $(n+k)!\le (2n)!$ for each nonegative integer $k\le n$. Thus

$$|T_n|\le \sum_{k=0}^n \ln^{-k-1} \pi \frac{(2n)!}{(n-k)!k!}=$$ $$\frac{(2n)!}{n!}\sum_{k=0}^n \ln^{-k-1} \pi \frac{n!}{(n-k)!k!}=\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$

Unfortunately, I expect that we cannot compensate the growth of this bound by suitable choice of $f(n)$. Indeed, approximating $|S_n-\pi|$ by the last summand $$\frac{4}{16^k(8k+1)} -\frac{2}{16^k(8k+4)} -\frac{1}{16^k(8k+5)} -\frac{1}{16^k(8k+6)}$$ in the sum for $S_n$, we obtain that $$|S_n-\pi|\approx \frac{120k^2+151k+47}{(8k+1)(8k+4)(8k+5)(8k+6)}\approx \frac{15}{64}16^{-k}k^{-2},$$ where $k=f(n)$. This value should compensate the value $$\frac{(2n)!}{n!\ln\pi}\left(1+\frac 1{\ln\pi}\right)^n.$$ Neglecting the factors of the smaller growth, we obtain that $16^{f(n)}\ge\approx \frac{(2n)!}{n!}$. But then $$\lim_{n\to\infty} \frac{16^{f(n)}e^{8f(n)}}{n!}\ge\approx e^{8f(n)}\frac{(2n!)}{n!^2}=\lim_{n\to\infty} e^{8f(n)}{2n\choose n}=\infty,$$ which violates $(2)$.